definition
Centre of Mass
The
center of mass of a body or a system of bodies is a mean position
of the total weight of the body where the resultant of the
forces applied is considered to be acted upon such that forces, momentum
and energy are conserved. The body or system of bodies is balanced
around the center of mass and the average of the weighted position
coordinates defines its coordinates.
2
formula
Centre of Mass Formula
Formula for centre of mass is:
X=m1x1+m2x2........+mnxnm!+m2+.....+mn=ΣmixiΣmi
X=m1x1+m2x2........+mnxnm!+m2+.....+mn=ΣmixiΣmi
3
definition
Linear, Area & Volume Mass Density
The
area density of a two-dimensional object is calculated as the mass per
unit area. The SI derived unit is: kilogram per square metre.
The linear density of a two-dimensional object is calculated as the mass per unit length. The SI derived unit is: kilogram per metre.
The volume density of a two-dimensional object is calculated as the mass per unit volume. The SI derived unit is: kilogram per metre cube.
The linear density of a two-dimensional object is calculated as the mass per unit length. The SI derived unit is: kilogram per metre.
The volume density of a two-dimensional object is calculated as the mass per unit volume. The SI derived unit is: kilogram per metre cube.
4
definition
Center of mass of two point masses
Center of mass of two point masses is given by:
→rCOM=m1→r1+m2→r2m1+m2
Example:
In a carbon monoxide molecule, the carbon and the oxygen atoms are separated by a distance 1.2×10−10m. Then find the radius of carbon atom is (Assume both atoms touch each other).Solution:
If centre of mass is taken as origin, then r1r2=m2m1
⇒r1r1+r2=m2m1+m2
⇒r1=1612+16×1.2×10−10=0.68×10−10m
→rCOM=m1→r1+m2→r2m1+m2
Example:
In a carbon monoxide molecule, the carbon and the oxygen atoms are separated by a distance 1.2×10−10m. Then find the radius of carbon atom is (Assume both atoms touch each other).Solution:
If centre of mass is taken as origin, then r1r2=m2m1
⇒r1r1+r2=m2m1+m2
⇒r1=1612+16×1.2×10−10=0.68×10−10m
5
definition
Center of mass of more than two point masses
Center of mass for a multiple mass system is given by:
→rCOM=∑→rimi∑mi,i∈1,2,3,...,n
Example:
Four particles are in x−y plane at
(1)1kg at (0,0)(2)2kg at (1,0)(3)3kg at (1,2)(4)4kg at (2,0)
Find where the center of mass is located.Solution:
Centre of mass is given by the relation, (∑miXi)/∑mi and (∑miYi)/∑miThus we get ((1(0)+2(1)+4(2)+3(1))(1+2+3+4),1(0)+2(0)+3(2)+4(0)1+2+3+4)
Thus we get the coordinates as (1.3,0.6)
→rCOM=∑→rimi∑mi,i∈1,2,3,...,n
Example:
Four particles are in x−y plane at
(1)1kg at (0,0)(2)2kg at (1,0)(3)3kg at (1,2)(4)4kg at (2,0)
Find where the center of mass is located.Solution:
Centre of mass is given by the relation, (∑miXi)/∑mi and (∑miYi)/∑miThus we get ((1(0)+2(1)+4(2)+3(1))(1+2+3+4),1(0)+2(0)+3(2)+4(0)1+2+3+4)
Thus we get the coordinates as (1.3,0.6)
6
definition
Use principle of symmetry to find centre of mass of system of discrete masses
For a discrete set of particles numbered 1,2,3,4,........,i,.........N each of mass m1,m2,m3,.....mi,.....,mN at positions (x1,y1,z1),(x2,y2,z2),.....(xi,yi,zi),......(xN,yN,zN). Let the total mass of the system be M.
The x-coordinate of centre of mass of the system is given by:
X=∑mixiNi=1M
The y-coordinate of centre of mass of the system is given by:
Y=∑miyiNi=1M
The z-coordinate of centre of mass of the system is given by:
Z=∑miziNi=1M
The x-coordinate of centre of mass of the system is given by:
X=∑mixiNi=1M
The y-coordinate of centre of mass of the system is given by:
Y=∑miyiNi=1M
The z-coordinate of centre of mass of the system is given by:
Z=∑miziNi=1M
7
definition
Centre of mass of Continuous Bodies
Centre
of mass of bodies coincide with their geometric centres and this this
can be determined by method of symmetry. For instance centre of mass of a
uniform semicircular disc lies on the vertical axis as the object is
symmetric about this.
Formula for finding centre of mass of continuous system:
Xcm=∫rdm∫dm
Formula for finding centre of mass of continuous system:
Xcm=∫rdm∫dm
8
example
Centre of Mass of continuous one dimensional bodies
Example: A non-uniform rod having mass per unit length as μ=ax (a is constant). If its total mass is M and length L.Find position of the centre of mass.
Solution:
Choose a coordinate system with the rod aligned along the x-axis and origin located at the left end of the rod. Choose an infinitesimal mass
element dm located a distance x'. Let the length of the mass element be dx'.
Thus
dm=μ(x′)dx′
The total mass is found by integrating the mass element over the length of the rod
M=∫L0μ(x′)dx′=a∫L0x′dx′=a2x′2∣L0=a2L2
or
a=2ML2
Now center of mass is calculated as
xcm=1M∫bodyxdm=1M∫L0μ(x′)x′dx′=aM∫L0x′2dx′
substituting the value of a
2L2∫L0x′2dx′=23L2x′3∣L0=23L2(L3−0)=23L
Solution:
Choose a coordinate system with the rod aligned along the x-axis and origin located at the left end of the rod. Choose an infinitesimal mass
element dm located a distance x'. Let the length of the mass element be dx'.
Thus
dm=μ(x′)dx′
The total mass is found by integrating the mass element over the length of the rod
M=∫L0μ(x′)dx′=a∫L0x′dx′=a2x′2∣L0=a2L2
or
a=2ML2
Now center of mass is calculated as
xcm=1M∫bodyxdm=1M∫L0μ(x′)x′dx′=aM∫L0x′2dx′
substituting the value of a
2L2∫L0x′2dx′=23L2x′3∣L0=23L2(L3−0)=23L
9
definition
Centre of Mass of a Half Disc
Center of Mass of a Half a Disk
→rc.m.=(xc.m.,yc.m.)=(0,2rπ)
yc.m.=∫dmMyring=∫dmM2rπ
Surface Density : The mass per unit area
σ=dmdA⟶dm=σdA
dA=π(r+dr)2−π(r)2
dA=πr2+2πrdr+π(dr)2−πr2
dA=2πrdr+π(dr)2
dA=2πrdr
dm=σdA=σπrdr=MAtotπrdr=M12πR2πrdr
dm=MR22rdr
→rc.m.=(xc.m.,yc.m.)=(0,2rπ)
yc.m.=∫dmMyring=∫dmM2rπ
Surface Density : The mass per unit area
σ=dmdA⟶dm=σdA
dA=π(r+dr)2−π(r)2
dA=πr2+2πrdr+π(dr)2−πr2
dA=2πrdr+π(dr)2
dA=2πrdr
dm=σdA=σπrdr=MAtotπrdr=M12πR2πrdr
dm=MR22rdr
10
definition
Center of mass of continuous three dimensional objects
Center of mass of continuous three dimensional objects is found by:
→rCOM=∫→rdm∫dm
Example:
Distance of the center of mass of a solid uniform cone from its vertex is z0 . If the radius of its base is R and its height is h then find z0.
Solution:
Suppose the cylindrical symmetry of the problem to note that the center of mass must lie along the z-axis (x=y=0). The only issue is how high does it lie. If the uniform density of the cone is ρ , then first compute the mass of the cone. If we slice the cone into circular disks of area πr2 and height dz, the mass is given by the integral:
M=∫ρdV=ρh∫0πr2dz
However, we know that the radius r starts at a for z=0, and goes linearly to zero when z=h. This means that r=a(1−z/h), so that:M=ρh∫0πa2(1−z/h)2dz=πa2ρh∫0(1−2z/h+z2/h2)dz=1/3πa2hρ
Now this simply indicates that the volume of the cone is given by
V=13πa2h
To find the height of the center of mass, we then compute: zcm=1M∫ρzdV=ρMh∫0πr2zdz=πa2ρMh∫0(1−zh)2zdz =πa2ρMh∫0(z−2z2h+z3h)dz=112Mπa2h2ρ=14h Note:
Choice of correct coordinate system and correct elemental sections is very important to make calculations easier.
→rCOM=∫→rdm∫dm
Example:
Distance of the center of mass of a solid uniform cone from its vertex is z0 . If the radius of its base is R and its height is h then find z0.
Solution:
Suppose the cylindrical symmetry of the problem to note that the center of mass must lie along the z-axis (x=y=0). The only issue is how high does it lie. If the uniform density of the cone is ρ , then first compute the mass of the cone. If we slice the cone into circular disks of area πr2 and height dz, the mass is given by the integral:
M=∫ρdV=ρh∫0πr2dz
However, we know that the radius r starts at a for z=0, and goes linearly to zero when z=h. This means that r=a(1−z/h), so that:M=ρh∫0πa2(1−z/h)2dz=πa2ρh∫0(1−2z/h+z2/h2)dz=1/3πa2hρ
Now this simply indicates that the volume of the cone is given by
V=13πa2h
To find the height of the center of mass, we then compute: zcm=1M∫ρzdV=ρMh∫0πr2zdz=πa2ρMh∫0(1−zh)2zdz =πa2ρMh∫0(z−2z2h+z3h)dz=112Mπa2h2ρ=14h Note:
Choice of correct coordinate system and correct elemental sections is very important to make calculations easier.
11
example
Principle of Negative mass to find centre of mass for a discrete mass system
Example: Look at the drawing given in the figure which has been drawn with ink
of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink used to draw the outer circle is 6 m. The coordinates of the centres of the different parts are: outer circle (0, 0) left inner circle (a, a), right inner circle (a, a),vertical line (0, 0) and horizontal line (0, a). The y-coordinate of the centre of mass of the ink in the above drawing is:
Solution: YCM=6m×0+ma+ma+−ma+m×06m+m+m+m+m
YCM=ma10m=a10
of uniform line-thickness. The mass of ink used to draw each of the two inner circles, and each of the two line segments is m. The mass of the ink used to draw the outer circle is 6 m. The coordinates of the centres of the different parts are: outer circle (0, 0) left inner circle (a, a), right inner circle (a, a),vertical line (0, 0) and horizontal line (0, a). The y-coordinate of the centre of mass of the ink in the above drawing is:
Solution: YCM=6m×0+ma+ma+−ma+m×06m+m+m+m+m
YCM=ma10m=a10
12
example
Principle of Negative mass to find centre of mass
Example: In the figure shown, find out the distance of centre of mass of a system of a uniform circular plate of radius 3R from O. On this plate, a hole
of radius R is cut whose centre is at 2R distance from centre of the
plate.
Solution: From law of conservation of momentum,
ˉx=m1x1+(−m2)x2m1+(−m2)
=A1x1+(−A2)x2A1+(−A2)
A1=π(3R)2,
A2=πR2
x1=0,
x2=2R
∴ˉx=−R/4
of radius R is cut whose centre is at 2R distance from centre of the
plate.
Solution: From law of conservation of momentum,
ˉx=m1x1+(−m2)x2m1+(−m2)
=A1x1+(−A2)x2A1+(−A2)
A1=π(3R)2,
A2=πR2
x1=0,
x2=2R
∴ˉx=−R/4
13
definition
Velocity of Centre of Mass
Vcm=m1v1+m2v2+...m1+m2+...
The velocity of centre of mass for a system of particles is defined this way and given by above formula.
The velocity of centre of mass for a system of particles is defined this way and given by above formula.
14
example
Acceleration of Centre of Mass
Acceleration of centre of mass is defined by:
acm=m1a1+m2a2+...mnanm1+m2+...mn
acm=m1a1+m2a2+...mnanm1+m2+...mn
15
example
Example on change in position of centre of mass
Example:
A mass m is at rest on a inclined plane structure of mass M which is
further resting on a smooth horizontal plane. Now if the mass starts
moving, Find how the position of centre of mass of the system will
change.
Solution:
Here the system is wedge+block. Net force on the system in horizontal direction is "0". Hence the centre of mass of the system will not move in horizontal direction. Now for vertical direction there is a force that is due to the mass of the wedge and the block and hence the centre of mass changes in this direction.
Solution:
Here the system is wedge+block. Net force on the system in horizontal direction is "0". Hence the centre of mass of the system will not move in horizontal direction. Now for vertical direction there is a force that is due to the mass of the wedge and the block and hence the centre of mass changes in this direction.
16
example
Position of centre of mass of two blocks attached in spring
Example: An elastic spring is compressed between two blocks of masses 1 kg and 2 kg resting on a smooth horizontal table as shown. If the spring has 12 J of energy and suddenly released, the velocity with which the larger block of 2 kg moves will be:
Solution: Using momentum conervation
mA.vA=mBvB
⇒vA=2vB
Now total K.E=P.E of spring
12mAv2A+12mB(vB)2=12
12mB2×4v2B+12mB(vB)2=12
3(vB)2=12
vB=2 m/s
Solution: Using momentum conervation
mA.vA=mBvB
⇒vA=2vB
Now total K.E=P.E of spring
12mAv2A+12mB(vB)2=12
12mB2×4v2B+12mB(vB)2=12
3(vB)2=12
vB=2 m/s
17
example
Newton's Law of motion to system of masses
Example: Two blocks A and B of mass m and 2m are connected together by a light spring of stiffness k. The system is lying on a smooth horizontal surface with the block A in contact with a fixed vertical wall as shown in the figure. The block B is pressed towards the wall by a distance x0 and then released. There is no friction anywhere. If spring takes time Δt to acquire its natural length then what is the average force on the block A by the wall.
Solution: Let v is the velocity of mass 2m in natural length of spring, then from conservation of energy we have,
12kx20=12(2m)v2
∴v=√k2mx0
Velocity of centre of mass at this instant,
vcm=Total momentumTotal mass
vcm=2mv3m
v=13√2kmx0
Now, impulse on system = change in momentum of system
∴FavΔt=(3m)vCM=(√2mk)x0
∴Fav=(√2mk)x0Δt
Solution: Let v is the velocity of mass 2m in natural length of spring, then from conservation of energy we have,
12kx20=12(2m)v2
∴v=√k2mx0
Velocity of centre of mass at this instant,
vcm=Total momentumTotal mass
vcm=2mv3m
v=13√2kmx0
Now, impulse on system = change in momentum of system
∴FavΔt=(3m)vCM=(√2mk)x0
∴Fav=(√2mk)x0Δt
18
definition
Linear Momentum
Linear momentum is a vector quantity defined as the product of an object's mass, m, and its velocity, v. Linear momentum is denoted by the letter p
and is called momentum for short. Note that a body's momentum is always
in the same direction as its velocity vector. The unit of momentum are
kg. m/s.
→p=m→v
→p=m→v
19
example
Explanation of Physical Phenomena using principle of conservation of Momentum
Example: Two balls of equal masses are thrown upwards along the same vertical line at an interval of 2 seconds with the same initial velocity of 39.2ms−1. Find the total time of flight of each ball, if they collide at a certain height, and the collision is perfectly inelastic.
Solution:
When the particles collide the height of both particles will be same,
For first particle time will be t and for second particle time will be t−2
h=ut−12gt2=u(t−2)−12g(t−2)2
ut−12gt2=ut−2u−12g(t2−4t+4)
0=−2u+2gt−2g
t=ug+1
t=5s
By solving the quadratic equation in t we get the time of first collision to be 5s.
h=39.2×5−12g×52=20g−12.5g=7.5g meter
Now by momentum conservation, velocity of particles after collision
m(u−gt)+m(u−g(t−2))=mv′
mu−5mg+mu−3mg=mv′
v′=2u−8g=0m/s
a=−g
Now for collision with ground let time taken be t2
h=12gt22⇒t=√2hg=√2×7.5gg=√15
Thus, times of total time of flights will be 5+√15s and 3+√15s
Solution:
When the particles collide the height of both particles will be same,
For first particle time will be t and for second particle time will be t−2
h=ut−12gt2=u(t−2)−12g(t−2)2
ut−12gt2=ut−2u−12g(t2−4t+4)
0=−2u+2gt−2g
t=ug+1
t=5s
By solving the quadratic equation in t we get the time of first collision to be 5s.
h=39.2×5−12g×52=20g−12.5g=7.5g meter
Now by momentum conservation, velocity of particles after collision
m(u−gt)+m(u−g(t−2))=mv′
mu−5mg+mu−3mg=mv′
v′=2u−8g=0m/s
a=−g
Now for collision with ground let time taken be t2
h=12gt22⇒t=√2hg=√2×7.5gg=√15
Thus, times of total time of flights will be 5+√15s and 3+√15s
20
example
Example on momentum conservation in two blocks attached by a spring
Two blocks of masses m and 3m are placed on a friction less, horizontal surface. A light spring is attached to the more massive block, and the
blocks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after which the block of mass 3m moves to the right with a speed of →V3m=2.60→i m/s.)What is the velocity of the block of mass m? (Assume right is positive and left is negative.)
Momentum is conserved and we have :
m→vm+3m→v3m=→0
→vm=−3mm→v3m
=−3→v3m
=−3×2.60→i
=−7.80→i
|→vm|=7.80m/s
blocks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after which the block of mass 3m moves to the right with a speed of →V3m=2.60→i m/s.)What is the velocity of the block of mass m? (Assume right is positive and left is negative.)
Momentum is conserved and we have :
m→vm+3m→v3m=→0
→vm=−3mm→v3m
=−3→v3m
=−3×2.60→i
=−7.80→i
|→vm|=7.80m/s
21
example
Problems on phenomenon related to Rocket Propulsion
Example: Like
rocket propulsion problem of variable mass can think of rain drops
falling through a cloud of small water droplets. Some of these small
droplets adhere to the rain drop. Thereby increasing its mass as it
falls. The force on the rain drops is
Fext=dpdt=mdvdt+vdmdt
Assume the mass of the raindrop depends on the distance x that it has fallen. Then m=kx, where k is a constant and
dmdt=kv, This gives since Fext=mg
mg=mdvdt+v(kv)
dividing by k, we get xg=xdvdt+v2 where x=mk
This is a differential equation that has a solution of the form v=at, where a is acceleration and is constant. Take initial velocity of the raindrops to zero. Then find the value of a.
Solution:
xdvdt+v2=gx may be written as dvdt+v2x=g
Its solution is ve∫dtt=∫ge∫dttdt
Solution to this differential equation is:
vt=gt22
v=at we get a=g2
Fext=dpdt=mdvdt+vdmdt
Assume the mass of the raindrop depends on the distance x that it has fallen. Then m=kx, where k is a constant and
dmdt=kv, This gives since Fext=mg
mg=mdvdt+v(kv)
dividing by k, we get xg=xdvdt+v2 where x=mk
This is a differential equation that has a solution of the form v=at, where a is acceleration and is constant. Take initial velocity of the raindrops to zero. Then find the value of a.
Solution:
xdvdt+v2=gx may be written as dvdt+v2x=g
Its solution is ve∫dtt=∫ge∫dttdt
Solution to this differential equation is:
vt=gt22
v=at we get a=g2
22
example
Problems on motion in two dimension where momentum is conserved in one dimension
Example: Two small identical spheres each of mass m are projected slant-wise from the points A and B on the ground with equal velocities u and making same angles θ and θ as shown in the figure. They collide with each other at the highest point C
of the common path. If the collision is completely inelastic, how much
time after the collision the particles come back to the ground?
Solution:
At the highest point, the velocity of the masses will only be horizontal and will be equal to ucosθ.
After the masses stick together, from the principle of
conservation of momentum:
v=mucosθ−mucosθm+m=0
So the body will be in free fall starting from rest.
For the vertical motion
12gt2=Hmax=u2sin2θ2g
Solving this, we get t=usinθg
Solution:
At the highest point, the velocity of the masses will only be horizontal and will be equal to ucosθ.
After the masses stick together, from the principle of
conservation of momentum:
v=mucosθ−mucosθm+m=0
So the body will be in free fall starting from rest.
For the vertical motion
12gt2=Hmax=u2sin2θ2g
Solving this, we get t=usinθg
23
example
Mass in motion separates in into multiple masses
Example: At a certain height a shell at rest explodes into two equal fragments. One of the fragments receives a horizontal velocity u. The time interval after which, the velocity vectors will be inclined at 120o to each other is:
Solution: ∵ shell explodes in equal fragments, their masses will be equal,
by conservation of momentum
2m(0)=mu+mu2
∴u2=−u
for angle to be 120∘
v=v1cos60∘
vn=u=v1sin60∘⇒v1=usin60∘
v=u+at
v1cos60∘=0+(+g)t⇒usin60∘cos60∘=gt
⇒ug1/2√3/2=t⇒t=u√3⋅g
Solution: ∵ shell explodes in equal fragments, their masses will be equal,
by conservation of momentum
2m(0)=mu+mu2
∴u2=−u
for angle to be 120∘
v=v1cos60∘
vn=u=v1sin60∘⇒v1=usin60∘
v=u+at
v1cos60∘=0+(+g)t⇒usin60∘cos60∘=gt
⇒ug1/2√3/2=t⇒t=u√3⋅g
24
example
Problem on Projectile and its explosion
Question: A projectile is moving at 20 m/s at its highest point, where it breaks into equal parts due to an internal explosion. One part moves vertically up at 30 m/s with respect to the ground. Then the other part will move with what velocity?
Solution:
Using momentum conservation,
m×20ˆi=m/2×30ˆj+m/2×x
40ˆi−30ˆj=x⇒x=50 m/s.
Solution:
Using momentum conservation,
m×20ˆi=m/2×30ˆj+m/2×x
40ˆi−30ˆj=x⇒x=50 m/s.
25
definition
Thrust Force
Example: Sand drop from a stationary hopper at a rate dmdt on to a conveyor belt moving with a velocity v in the reference frame of the laboratory. Since dvdt=0,
Therefore, it is an example of variable mass. To an observer at rest on
the belt the falling sand would appear to have a horizontal motion with
speed v in a direction opposite to that shown for the belt in the laboratory. Therefore vrel=−v. In this example dmdt
is positive as the system is gaining mass. Note that in the absence of
friction mass of the belt does not enter the problem. power supplied by
the external force is P=¯F⋅¯v=¯v⋅(¯vdmdt)=v2dmdt.
Solution:
Since KE=12mv2 and d(KE)dt=v22dmdt.
Also, P=v2dmdt=2d(KE)dt
Solution:
Since KE=12mv2 and d(KE)dt=v22dmdt.
Also, P=v2dmdt=2d(KE)dt
26
example
Problems on variable mass system
Example:
Two blocks of masses m1 and m2 are connected by spring of constant K. The spring is initially compressed and the system is released from rest at t = 0 second. The work done by spring on the blocks m1 and m2 be W1 and W2 respectively by time t. The speeds of both the blocks at time t are non zero. Then find the value of W1W2 .
Solution:
Let work done by spring on the block of mass m1 be W1= change in kinetic energy
=12m1v12−12m1(0)2=12m1v12
Similarly, W2=12m2v22
⇒W1W2=12m1v2112m2v22
From conservation of momentum,
m1v1=m2v2
∴W1W2=v1v2=m2m1
Two blocks of masses m1 and m2 are connected by spring of constant K. The spring is initially compressed and the system is released from rest at t = 0 second. The work done by spring on the blocks m1 and m2 be W1 and W2 respectively by time t. The speeds of both the blocks at time t are non zero. Then find the value of W1W2 .
Solution:
Let work done by spring on the block of mass m1 be W1= change in kinetic energy
=12m1v12−12m1(0)2=12m1v12
Similarly, W2=12m2v22
⇒W1W2=12m1v2112m2v22
From conservation of momentum,
m1v1=m2v2
∴W1W2=v1v2=m2m1
27
example
Momentum conservation of two blocks attached by a spring
Example: A boat of length 10 m and mass 450 kg is floating without motion in still water. A man of mass 50 kg
standing at one end of it walks to the other end of it and stops. The
magnitude of the displacement of the boat in meters relative to ground
is:
Solution:
Centre of mass will not be displaced, thus
m1x1+m2x2=0
Now, let x1=x,thenx2=−(10−x)
(because distance covered by man relative to boat = 10 m or
x2−(−x1)=10,(in opposite direction)
Thus, 450x−50(10−x)=0orx=1 m
Solution:
Centre of mass will not be displaced, thus
m1x1+m2x2=0
Now, let x1=x,thenx2=−(10−x)
(because distance covered by man relative to boat = 10 m or
x2−(−x1)=10,(in opposite direction)
Thus, 450x−50(10−x)=0orx=1 m
28
example
Problem based on conservation of momentum in elastic collision
Example: A ball of mass M moving with a velocity V collides head on elastically with another of same mass but moving with a velocity v in the opposite direction. What will be the scenario after collision?
Solution:
Using conservation of momentum we have
M1u1−M2u2=−M1v1+M2v2or
MV−Mu2=−Mv+Mv2
or
V−u2=−v+v2.............(i)
and as for a perfect elastic collision we have e=1, thus we get
v2+v=u2+V.................(ii)
Using these two equations we get
u2=v and v2=V
Thus the velocities are exchanged between the two balls.
Solution:
Using conservation of momentum we have
M1u1−M2u2=−M1v1+M2v2or
MV−Mu2=−Mv+Mv2
or
V−u2=−v+v2.............(i)
and as for a perfect elastic collision we have e=1, thus we get
v2+v=u2+V.................(ii)
Using these two equations we get
u2=v and v2=V
Thus the velocities are exchanged between the two balls.
29
example
Conservation of kinetic energy in elastic collision
Example: A plank of mass M is placed on a smooth horizontal surface. Two light identical springs, each of stiffness K,
are rigidly connected to struts at the end of the plank as shown in the
figure. When the spring is in their unextended position, the distance
between their free ends is 3l. A block of mass m is placed on the plank and pressed against one of the springs so that it is compressed to l.
To keep the block at rest it is connected to the strut by means of a
light string. Initially, the system is at rest. Now the string is burnt.
Find the maximum kinetic energy.
Solution:
Using Conservation of energy and conservation of momentum, the equations we get are:
Kl22=12mv2+12MV2andmv+MV=0
where v= max velocity of m
and V= max velocity of M
Solving these two equations:
we get:
v=√KMm(M+m)l
KEmax=mv22=KMl22(M+m)
Solution:
Using Conservation of energy and conservation of momentum, the equations we get are:
Kl22=12mv2+12MV2andmv+MV=0
where v= max velocity of m
and V= max velocity of M
Solving these two equations:
we get:
v=√KMm(M+m)l
KEmax=mv22=KMl22(M+m)
30
example
Conservation of Energy during collision
Example: A ball of mass m moving with speed v undergoes a head-on elastic collision with a ball of mass nm initially at rest. What is the fraction of the incident energy transferred to the second ball?
Solution:Given :
m1=m
m2=nmLet the velocities of A and B after the collision be v1 and v2 respectively.initial velocity of A before collision is v and that of B is zero.Initial kinetic energy of the system E=12mv2
Using v2=(m2−em1)u2+(1+e)m1u1m1+m2 v2=(nm−1×m)(0)+(1+1)m(v)m+nm ⟹v2=2v(1+n)Thus Kinetic energy of B after the collision E′B=12(nm)v22
E′B=12(nm)4v2(1+n)2
Fraction of total kinetic energy retained by B E′BE=4n(n+1)2
Solution:Given :
m1=m
m2=nmLet the velocities of A and B after the collision be v1 and v2 respectively.initial velocity of A before collision is v and that of B is zero.Initial kinetic energy of the system E=12mv2
Using v2=(m2−em1)u2+(1+e)m1u1m1+m2 v2=(nm−1×m)(0)+(1+1)m(v)m+nm ⟹v2=2v(1+n)Thus Kinetic energy of B after the collision E′B=12(nm)v22
E′B=12(nm)4v2(1+n)2
Fraction of total kinetic energy retained by B E′BE=4n(n+1)2
31
example
Velocity of approach and velocity of separation for particles colliding in one dimension
Example:
A metal ball falls from a height of 1 m on to a steel plate and
jumps upto a height of 81 cm. Find the coefficient of restitution of the
ball material.
Solution: Let
u1=Velocity of ball before collision
v1=Velocity of ball after collision
u2=Velocity of plate before collision=0
v2=Velocity of plate after collision=0
The coefficient of restitution(e) is given by,
e=−Velocity of separationVelocity of approach
Velocity of separation = v1−v2
Velocity of approach = u1−u2
e=−v1−v2u1−u2
⇒e=−−v1−0u1−0=v1u1
Using v21=2gh1⇒v1=√2gh1
Using u21=2gh2⇒u1=√2gh2
⇒e=√2gh2√2gh1=√h2h1
Here, h1=100cm,h2=81cm
e=√h2h1=√81100=0.9
Solution: Let
u1=Velocity of ball before collision
v1=Velocity of ball after collision
u2=Velocity of plate before collision=0
v2=Velocity of plate after collision=0
The coefficient of restitution(e) is given by,
e=−Velocity of separationVelocity of approach
Velocity of separation = v1−v2
Velocity of approach = u1−u2
e=−v1−v2u1−u2
⇒e=−−v1−0u1−0=v1u1
Using v21=2gh1⇒v1=√2gh1
Using u21=2gh2⇒u1=√2gh2
⇒e=√2gh2√2gh1=√h2h1
Here, h1=100cm,h2=81cm
e=√h2h1=√81100=0.9
32
example
One dimensional elastic collision for bodies with equal masses
Question: Two elastic bodies P and Q having equal masses are moving along the same line with velocities of 16 m/s and 10 m/s respectively. What will be their velocities after the elastic collision will be (in m/s)?
Solution:
Let velocities be v1 and v2
conservation of momentum.
m(16+10)=mv1+mv2
⇒v1+v2=26
e=1.
16−10=v2−v1
The velocities are 10 m/s and 16 m/s
Solution:
Let velocities be v1 and v2
conservation of momentum.
m(16+10)=mv1+mv2
⇒v1+v2=26
e=1.
16−10=v2−v1
The velocities are 10 m/s and 16 m/s
33
example
Collision of masses in one dimension
Example: Two
identical balls A and B each of mass "m" are moving towards each other
at equal speed each of "v". Ball A initially is moving along the
positive x direction while ball B is moving along negative x direction.
If the collision is perfectly elastic, what will be the impulse received
by the ball B?
Solution:Since the collision is perfectly elastic, ball B travels in the backward direction i.e along positive x-axis with the same speed before collision.
The change in velocity becomes v−(−v)=2v
∴change in momentum=m(2v)=2mv
Which is the impulse received by the body.
Solution:Since the collision is perfectly elastic, ball B travels in the backward direction i.e along positive x-axis with the same speed before collision.
The change in velocity becomes v−(−v)=2v
∴change in momentum=m(2v)=2mv
Which is the impulse received by the body.
34
example
Problems on Multiple Collisions in one dimension
Example: Two balls of same mass each m are moving with same velocities v
on a smooth surface as shown in figure. If all collisions between the
mass and with the wall are perfectly elastic then what is possible
number of collisions between the bodies and wall together?
Solution: In case of elastic collision the velocity after the collision gets interchanged if masses of colliding bodies are same and the direction of velocity reverses if the one of the mass is very large compared to smaller mass and is at rest.
Using the above concepts the first collision will take place between wall and first ball, the velocity of ball will get reversed. Now the second collision will be between first and second ball where the of velocity will get interchanged now
the first ball will move again towards wall and second will move away from wall.
Now the third collision will take place between wall and ball moving towards it, the velocity of ball will get reversed away from wall. and both balls will move in same direction with constant velocity. Total there will be three collisions.
Solution: In case of elastic collision the velocity after the collision gets interchanged if masses of colliding bodies are same and the direction of velocity reverses if the one of the mass is very large compared to smaller mass and is at rest.
Using the above concepts the first collision will take place between wall and first ball, the velocity of ball will get reversed. Now the second collision will be between first and second ball where the of velocity will get interchanged now
the first ball will move again towards wall and second will move away from wall.
Now the third collision will take place between wall and ball moving towards it, the velocity of ball will get reversed away from wall. and both balls will move in same direction with constant velocity. Total there will be three collisions.
35
definition
Elastic collision where one of the mass is very large
Let two masses colliding have mass m1,m2, initial velocities u1,u2 and final velocities v1,v2. Let the collision be in one-dimension. Setting u2=0 without loss of generality.
Then, using conservation of momentum,
m1v1+m2v2=m1u1
Using conservation of energy,
m1u21=m1v21+m2v22
Solving,
v1=m1−m2m1+m2u1
v2=2m1m1+m2u1
Now, since m2>>m1,
v1≈−u1
v2=0
Then, using conservation of momentum,
m1v1+m2v2=m1u1
Using conservation of energy,
m1u21=m1v21+m2v22
Solving,
v1=m1−m2m1+m2u1
v2=2m1m1+m2u1
Now, since m2>>m1,
v1≈−u1
v2=0
36
definition
Reduced Mass and Problem involving reduced mass
Definition:
When two bodies in relative motion are acted upon by a central force
involving Newton's law then the system can be replaced by a single mass
called the reduced mass.
Example: A light spring of force constant K is held between two blocks of masses m and 2m. The two blocks and the spring system rests on a smooth horizontal floor. Now the blocks are moved towards each other compressing the spring by x and then they are suddenly released. Then find the relative velocity between the blocks when the spring attains its natural length.
Solution:Potential energy of the spring is being converted into kinetic energy of system.
Potential energy of the spring before released =12Kx2
K.E of the system in Center of mass frame =12μv2relative , where μ=m×2mm+2m=2m3
is reduced mass of the system.
Now, 12μv2relative=12Kx2
v2relative=32mKx2
vrelative=(√3K2m)x
Example: A light spring of force constant K is held between two blocks of masses m and 2m. The two blocks and the spring system rests on a smooth horizontal floor. Now the blocks are moved towards each other compressing the spring by x and then they are suddenly released. Then find the relative velocity between the blocks when the spring attains its natural length.
Solution:Potential energy of the spring is being converted into kinetic energy of system.
Potential energy of the spring before released =12Kx2
K.E of the system in Center of mass frame =12μv2relative , where μ=m×2mm+2m=2m3
is reduced mass of the system.
Now, 12μv2relative=12Kx2
v2relative=32mKx2
vrelative=(√3K2m)x
37
example
Conservation of momentum in two dimensions
Considering
two balls as a system, there isn't any external force acting on them.
Momentum will be conserved in either directions along the motion and
perpendicular to it.
¯p1+¯p2=¯p′1+¯p′2
¯x→mv+0=¯p′1+¯p′2 ¯y→0+0=¯p′1+¯p′2 ¯p1x=m|v′1|cosθ
¯p2x=m|v′2|cosϕ
¯p1y=m|v′1|sinθ¯p1y=m|v′2|sinϕ12mv2=12mv′21+12mv′22
v2=v′21+v′22
¯p1+¯p2=¯p′1+¯p′2
¯x→mv+0=¯p′1+¯p′2 ¯y→0+0=¯p′1+¯p′2 ¯p1x=m|v′1|cosθ
¯p2x=m|v′2|cosϕ
¯p1y=m|v′1|sinθ¯p1y=m|v′2|sinϕ12mv2=12mv′21+12mv′22
v2=v′21+v′22
38
example
Conservation of energy in elastic collision
Example: A
ball A moving with a velocity 5 m/s collides elastically with
another identical ball at rest such that the velocity of A makes an
angle of 30∘ with the line joining the centres of the balls. Then what will be the scenario after collision?
Solution: As the collision is elastic so kinetic energy as well as momentum both are conserved.
For momentum conservation,
mv=mvAcos300+mvBcosθ⇒v=vAcos300+vBcosθ....(1)
0=mvAsin300−mvBsinθ⇒0=vAsin300−vBsinθ.....(2)
For KE conservation,
12mv2=12m(v2Acos2300+v2Bcos2θ)
v2=v2Acos2300+v2Bcos2θ....(3) and
0=v2Asin2300+v2Bsin2θ.....(4)
(3)+(4),v2=v2A+v2B....(5)
now squaring (1) and (2) and then adding, v2=v2A+v2B+vAvBcos(300+θ)
using (5), cos(300+θ)=0=cos900⇒θ=90−30=60o
Thus A and B will move at right angle after collision.
putting θ=600 in (1) and (2), v=vAcos300+vBcos600⇒5=√32vA+vB2...(6)
and 0=vA2−√32vB...(7)
solving (6) and (7), vA=5√32m/s and vB=52m/s
Solution: As the collision is elastic so kinetic energy as well as momentum both are conserved.
For momentum conservation,
mv=mvAcos300+mvBcosθ⇒v=vAcos300+vBcosθ....(1)
0=mvAsin300−mvBsinθ⇒0=vAsin300−vBsinθ.....(2)
For KE conservation,
12mv2=12m(v2Acos2300+v2Bcos2θ)
v2=v2Acos2300+v2Bcos2θ....(3) and
0=v2Asin2300+v2Bsin2θ.....(4)
(3)+(4),v2=v2A+v2B....(5)
now squaring (1) and (2) and then adding, v2=v2A+v2B+vAvBcos(300+θ)
using (5), cos(300+θ)=0=cos900⇒θ=90−30=60o
Thus A and B will move at right angle after collision.
putting θ=600 in (1) and (2), v=vAcos300+vBcos600⇒5=√32vA+vB2...(6)
and 0=vA2−√32vB...(7)
solving (6) and (7), vA=5√32m/s and vB=52m/s
39
definition
Velocity of approach and velocity of separation for colliding particles
40
example
Elastic collision of equal masses in two dimensions
Let a body of mass m collide with an object of same mass at rest. Let its velocity be un along the normal before collision and ut along the tangent.
Using conservation of momentum in tangential direction,
mut=mv1,t
v1,t=ut
Using conservation of momentum in normal direction,
mun=mv1,n+mv2,n
For elastic collision, velocity of approach equals the velocity of separation
v2,n−v1,n=un
Solving,
v1,n=m1−m2m1+m2un
v2,n=2m1m1+m2un
Using m2=m1,
v1,n=0
v2,n=un
Using conservation of momentum in tangential direction,
mut=mv1,t
v1,t=ut
Using conservation of momentum in normal direction,
mun=mv1,n+mv2,n
For elastic collision, velocity of approach equals the velocity of separation
v2,n−v1,n=un
Solving,
v1,n=m1−m2m1+m2un
v2,n=2m1m1+m2un
Using m2=m1,
v1,n=0
v2,n=un
41
example
Collision of masses in two dimensions
Example: Two small identical spheres each of mass m are projected slant-wise from the points A and B on the ground with equal velocitiesu and making same angles θ and θ as shown in the figure. They collide with each other at the highest point C
of the common path. If the collision is completely inelastic, how much
time after the collision the particles come back to the ground?
Solution: At the highest point, the velocity of the masses will only be horizontal and will be equal to ucosθ.
After the masses stick together, from the formula given in the concept
v=mucosθ−mucosθm+m=0
So the body will be in free fall starting from rest.
For the vertical motion
12gt2=Hmax=u2sin2θ2g
Solving this, we get t=usinθg
Solution: At the highest point, the velocity of the masses will only be horizontal and will be equal to ucosθ.
After the masses stick together, from the formula given in the concept
v=mucosθ−mucosθm+m=0
So the body will be in free fall starting from rest.
For the vertical motion
12gt2=Hmax=u2sin2θ2g
Solving this, we get t=usinθg
42
example
Application of multiple elastic collisions in two-dimensions
Example:
A ball is projected horizontally from top of a 80 m deep well with velocity 10 m/s. Then particle will fall on the bottom at a distance of (all the collisions with the wall are elastic)Solution:
Total time taken by the ball to reach at bottom =√2Hg=√2×8010=4 sec
Let time taken in one collision is tcollision. Then tcollision×10=7 since horizontal velocity is 10 m/s
⇒tcollision=0.7 sec.
No. of collisions =40.7=557 (5th collisions from wall B)
Horizontal distance travelled in between 2 successive collisions =7m
∴Horizontal distance travelled in 5/7 part of collisions = The 1st collision will be with wall B, 2nd cillision with wall A, 3rd collision with wall B, 4th collision with wall A and 5th collision will be with wall B. After 5th collision, the ball will cover a distance of 57×7=5m from wall B i.e it will have collision at a distance of 2 m from wall A.
A ball is projected horizontally from top of a 80 m deep well with velocity 10 m/s. Then particle will fall on the bottom at a distance of (all the collisions with the wall are elastic)Solution:
Total time taken by the ball to reach at bottom =√2Hg=√2×8010=4 sec
Let time taken in one collision is tcollision. Then tcollision×10=7 since horizontal velocity is 10 m/s
⇒tcollision=0.7 sec.
No. of collisions =40.7=557 (5th collisions from wall B)
Horizontal distance travelled in between 2 successive collisions =7m
∴Horizontal distance travelled in 5/7 part of collisions = The 1st collision will be with wall B, 2nd cillision with wall A, 3rd collision with wall B, 4th collision with wall A and 5th collision will be with wall B. After 5th collision, the ball will cover a distance of 57×7=5m from wall B i.e it will have collision at a distance of 2 m from wall A.
43
example
Problems on collision combined with splitting of a mass
Example: A particle is projected with a velocity of 20ms−1 at an angle
600 with the horizontal. At the maximum height it splits into two parts of equal masses. If one part just drops down, what is the velocity of the other part?
Solution:
Velocity at max height =20cos60
=10 m/s.
i.e.,
using momentum conservation
10×2m=m×v
⇒v=20 m/s.
600 with the horizontal. At the maximum height it splits into two parts of equal masses. If one part just drops down, what is the velocity of the other part?
Solution:
Velocity at max height =20cos60
=10 m/s.
i.e.,
using momentum conservation
10×2m=m×v
⇒v=20 m/s.
44
example
Elastic collision with infinite mass in two dimensions
Let a body of mass m1 collide with an infinite mass at rest. Let its velocity be un along the normal before collision and ut along the tangent.
Using conservation of momentum in tangential direction,
m1ut=m1v1,t
v1,t=ut
Using conservation of momentum in normal direction,
m1un=m1v1,n+m2v2,n
For elastic collision, velocity of approach equals the velocity of separation
v2,n−v1,n=un
Solving,
v1,n=m1−m2m1+m2un
v2,n=2m1m1+m2un
Using approximation m2>>m1,
v1,n=−u1,n
v2,n=0
Using conservation of momentum in tangential direction,
m1ut=m1v1,t
v1,t=ut
Using conservation of momentum in normal direction,
m1un=m1v1,n+m2v2,n
For elastic collision, velocity of approach equals the velocity of separation
v2,n−v1,n=un
Solving,
v1,n=m1−m2m1+m2un
v2,n=2m1m1+m2un
Using approximation m2>>m1,
v1,n=−u1,n
v2,n=0
45
definition
Define reduced mass and use in solving problems in two dimensions
In a collision with a coefficient of restitution e, the change in kinetic energy can be written as:
ΔK=12μv2rel(e2−1)
where vrel is the relative velocity of the bodies before collision and μ is the reduced mass of the system.
ΔK=12μv2rel(e2−1)
where vrel is the relative velocity of the bodies before collision and μ is the reduced mass of the system.
46
example
Conservation of Momentum in inelastic collision
Example: A body of mass 10 g moving with a velocity of 20ms−1 collides with a stationary mass of 90 g. The collision is perfectly inelastic.
Find the percentage loss of kinetic energy of the system.
Solution:
Applying conservation of momentum.
10×20=100×v
v=2 m/s
ΔK.E=12×10×(20)2−12×100×(2)2
=2000−200
=1800
% loss =180012×10×400×100
=18002000×100=90%
Find the percentage loss of kinetic energy of the system.
Solution:
Applying conservation of momentum.
10×20=100×v
v=2 m/s
ΔK.E=12×10×(20)2−12×100×(2)2
=2000−200
=1800
% loss =180012×10×400×100
=18002000×100=90%
47
example
Velocity of approach and velocity of separation for particles colliding for inelastic collision
When
the two bodies stick together after collision, then it is perfectly
inelastic collision and in this case, the coefficient of restitution e is equal to zero.
In perfectly inelastic collision, the two colliding bodies gets stuck together and move with a common velocity.For perfectly inelastic collision, e=0.
In perfectly inelastic collision, total momentum of the system remains conserved just before and just after the collision but some kinetic energy is lost during the collision.
In perfectly inelastic collision, the two colliding bodies gets stuck together and move with a common velocity.For perfectly inelastic collision, e=0.
In perfectly inelastic collision, total momentum of the system remains conserved just before and just after the collision but some kinetic energy is lost during the collision.
48
example
Coefficient of restitution
Definition: The
coefficient of restitution (COR) is a measure of the "restitution" of a
collision between two objects: how much of the kinetic energy remains
for the objects to rebound from one another vs. how much is lost as
heat, or work done deforming the objects.
e=Velocity of separationVelocity of Approach
e=Velocity of separationVelocity of Approach
49
example
Problem in Inelastic Collision
Example: Two bodies move towards each other and collide inelastically. The velocity of the first body before impact is 2 m/s and of the second is 4 m/s.The common velocity after collision is 1
m/s in the direction of the first body. How many times did the K.E. of
the first body exceed that of the second body before collision.
Solution:
using momentum conservation
m1×2−4m2=(m1+m2)
⇒m1=5m2
Ratio of K.E=12×m1×(2)212×m2×(4)2
=5m2×4m2×16
=1.25
Solution:
using momentum conservation
m1×2−4m2=(m1+m2)
⇒m1=5m2
Ratio of K.E=12×m1×(2)212×m2×(4)2
=5m2×4m2×16
=1.25
50
example
Inelastic Collision in two dimensions
Example: A particle of mass m is projected from the ground with an initial speed u0at an angle αwith the horizontal. At
the highest point of its trajectory, it makes a completely inelastic
collision with another identical particle, which was thrown vertically
upward from the ground with the same initial speed u0. What is the angle that the composite system makes with the horizontal immediately after the collision?
Solution:
Velocity of the particle performing projectile motion at highest point
=V1=V0cosα
Velocity particle thrown vertically upwards at the position of collision
=v22=u20−2gu20sin2α2g=u0cosα
So, from conservation of momentum
tanθ=mu0cosαmu0cosα=1
⇒θ =π/4
Solution:
Velocity of the particle performing projectile motion at highest point
=V1=V0cosα
Velocity particle thrown vertically upwards at the position of collision
=v22=u20−2gu20sin2α2g=u0cosα
So, from conservation of momentum
tanθ=mu0cosαmu0cosα=1
⇒θ =π/4
51
example
Solving Problem involving completely inelastic collision
Example: Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. These have masses m,2m and m,
respectively.The object A moves towards B with a speed 9 m/s and makes
an elastic collision with it. Thereafter, B makes completely inelastic
collision with C. All motions occur on the same straight line. Find the
final speed (in m/s) of the object C.
Solution: After 1st collision
mvA=mv′A+2mv′B
−1=v′B−v′A0−vA⇒v′B=6m/s
After the 2nd collision
2mv′B=(2m+m)VC⇒vC=23v′B⇒vC=4m/s
Solution: After 1st collision
mvA=mv′A+2mv′B
−1=v′B−v′A0−vA⇒v′B=6m/s
After the 2nd collision
2mv′B=(2m+m)VC⇒vC=23v′B⇒vC=4m/s
52
example
Problem on perfectly inelastic collision in one dimension
Example: A particle of mass m moving in the direction x with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, what is the percentage loss in the energy during the collision?
Solution:
as the collision is inelastic the masses will move together
Assuming the speed of block A and B becomes v1ˆi+v2ˆj
Writing momentum equation in x direction
m(2v)+2m(0)=3mv1⇒v1=2v3
Writing momentum equation in Y direction
m(0)+2m(v)=3mv2⇒v2=2v3
The velocity of both blocks will be 2v3ˆi+2v3ˆj
The loss in kinetic energy
12m(2v)2+122m(v)2−123m(√22v3)2
3mv2−43mv2=53mv2
percentage loss53mv23mv2×100=56 %
Solution:
as the collision is inelastic the masses will move together
Assuming the speed of block A and B becomes v1ˆi+v2ˆj
Writing momentum equation in x direction
m(2v)+2m(0)=3mv1⇒v1=2v3
Writing momentum equation in Y direction
m(0)+2m(v)=3mv2⇒v2=2v3
The velocity of both blocks will be 2v3ˆi+2v3ˆj
The loss in kinetic energy
12m(2v)2+122m(v)2−123m(√22v3)2
3mv2−43mv2=53mv2
percentage loss53mv23mv2×100=56 %
53
example
Problems on Change in Kinetic Energy
Example: A
body starts from rest and is acted on by a constant force. What is the
ratio of kinetic energy gained by it in the first five seconds to that
gained in the next five second?
Solution: 1) constant force → constant acceleration =a
⇒V5−v0t=a=V5−05=a⇒V5=5a
⇒K.E5=12mV25=252ma2
2) v10−v0t=a⇒v10=10a⇒K.E10=1002ma2
So K.E gained between 5s and 10s =K.E10−K.E5
=752ma2
K.E gained in 5s=252ma2
⇒Ratio=2575=1:3
Solution: 1) constant force → constant acceleration =a
⇒V5−v0t=a=V5−05=a⇒V5=5a
⇒K.E5=12mV25=252ma2
2) v10−v0t=a⇒v10=10a⇒K.E10=1002ma2
So K.E gained between 5s and 10s =K.E10−K.E5
=752ma2
K.E gained in 5s=252ma2
⇒Ratio=2575=1:3
54
example
Collisions combined with pulley mass system
Example:
P and Q are two identical masses at rest suspended by an inextensible string passing over a smooth frictionless pulley. Mass P is given a downward push with a speed v as shown in figure. It collides elastically with the floor and rebounds immediately. What happens immediately after collision?
Solution:
Just after collision, P reverses its direction of motion causing the string to become loose. This results in zero tension in the wire. Q continues to move upward with velocity v due to inertia.
P and Q are two identical masses at rest suspended by an inextensible string passing over a smooth frictionless pulley. Mass P is given a downward push with a speed v as shown in figure. It collides elastically with the floor and rebounds immediately. What happens immediately after collision?
Solution:
Just after collision, P reverses its direction of motion causing the string to become loose. This results in zero tension in the wire. Q continues to move upward with velocity v due to inertia.
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definition
Impulsive and Non-impulsive Force
The process of minimizing an impact force can be approached from the definition of the impulse of force:
If an impact stops a moving object, then the change in momentum is a fixed quantity, and extending the time of the collision will decrease the impact force by the same factor.
Non-impulsive Force: A constant force acting on a body is an example of non-impulsive force.
If an impact stops a moving object, then the change in momentum is a fixed quantity, and extending the time of the collision will decrease the impact force by the same factor.
Non-impulsive Force: A constant force acting on a body is an example of non-impulsive force.
56
example
Find impulsive force using impulse momentum relation
Example: A particle of mass 2 kg moving with a velocity of 3 m/s is acted upon by a force which changes its direction of motion by an angle of 90o without changing its speed. What is the magnitude of impulse experienced by the particle?
Solution:
Initial velocity of the particle
→v1=3ˆi m/s
Final velocity of the particle →v2=3ˆj m/s
Change in momentum Δ→P=m(→v2−→v1)=2×3(ˆj−ˆi)
|Δ→P|=2×3√2=6√2 N s
Magnitude of impulse I=|Δ→P|=6√2 N-s
Solution:
Initial velocity of the particle
→v1=3ˆi m/s
Final velocity of the particle →v2=3ˆj m/s
Change in momentum Δ→P=m(→v2−→v1)=2×3(ˆj−ˆi)
|Δ→P|=2×3√2=6√2 N s
Magnitude of impulse I=|Δ→P|=6√2 N-s
57
example
Collisions with pulley mass system with tension as impulsive force
Example:A block of mass m and a pan of the same mass are connected by a pulley as shown in the figure. A particle of mass m travelling with a velocity v collides and stick to the pan as shown. Find the velocity of blocks after collision if initially the pan is at rest.
Solution:
Let N be the contact force between the particle and the pan and V be the final velocity. Impulse imparted equals the change in momentum.
For particle, Ndt=mv−mV
For pan, (N−T)dt=mV
For block, Tdt=mV
Solving, V=v/3
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