MOTION IN A STRAIGHT LINE(LEVEL 2)

• 1.

  A goods train accelerating uniformly on a straight railway track, approaches an electric pole standing on the side of track. Its engine passes the pole with velocity u and the guard’s compartment passes with velocity v. The middle wagon of the train passes the pole with a velocity
  a 
  

  $\frac{u+v}{2}$
  b 
  

  $\frac{1}{2}\sqrt{u^2+{\text{v}}^2}$
  c 
  

  $\sqrt{u\text{v}}$
  d 
  

  $\sqrt{\left(\frac{u^2+{\text{v}}^2}{2}\right)}$
  Difficulty Meter: 
  

  Intense (Only less than 20% of users made it right)
  Explanation:

  Let 'S' be the distance between two ends and 'a' be the constant acceleration As we know v² – u² = 2aS

  or, aS = $\frac{{\text{v}}^2-u^2}{2}$

  Let v be velocity at mid point.

  Therefore, ${\text{v}}_c^2-u^2=2a\frac{S}{2}$

  ${\text{v}}_c^2=u^2+aS$

  ${\text{v}}_c^2=u^2+\frac{{\text{v}}^2-u^2}{2}$

  ${\text{v}}_c=\sqrt{\frac{u^2+{\text{v}}^2}{2}}$
  Subject: 
  

  Physics
  Correct Answer: d
  

  Comment · 0 · 
• 2.

  A ball is dropped on a floor and bounces back to a height somewhat less than the original height. Which of the curves depicts its motion correctly?
  a 
  

  
  b 
  

  
  c 
  

  
  d 
  

  
  Difficulty Meter: 
  

  Very Hard (About 20% - 40% of users made it right)
  Explanation:

  When a ball is dropped on a floor,
  $\text{y}=\frac{1}{2}g{t}^2.....\left(1\right)$
  So the graph between y and t is a parabola. Here as time increases, y decreases. When the ball bounces back, then
  $\text{y}=ut+\frac{1}{2}g{t}^2.....\left(2\right)$
  The graph between y and t will be a parabola. But here as time increases, y also increases. So (b) represents correct graph.
  Subject: 
  

  Physics
  Correct Answer: b
  

  Comment · 0 · 
• 3.

  A train of 150 metre length is going towards north direction at a speed of 10 m/s . A parrot flies at a speed of 5 m/s towards south direction parallel to the railway track. The time taken by the parrot to cross the train is
  a 
  

  12 sec
  b 
  

  8 sec
  c 
  

  15 sec
  d 
  

  10 sec
  Difficulty Meter: 
  

  Very Hard (About 20% - 40% of users made it right)
  Explanation:

  Relative velocity of parrot w.r.t the train
  = 10 – (–5) = 15 ms^–1.
  Time taken by parrot to cross the train
  $=\frac{distance}{speed}=\frac{150}{15}=10s$
  Subject: 
  

  Physics
  Correct Answer: d
  

  Comment · 0 · 
• 4.

  Three different objects of masses m₁ ,m₂ and m₃ are allowed to fall from rest from the same point along three different frictionless paths. The speeds of the three objects on reaching the ground will be in the ratio of
  a 
  

  m₁ : m₂ : m₃
  b 
  

  m₁ : 2m₂ : 3m₃
  c 
  

  1 : 1 : 1
  d 
  

  $\frac{1}{m_1}:\frac{1}{m_2}:\frac{1}{m_3}$
  Difficulty Meter: 
  

  Intense (Only less than 20% of users made it right)
  Explanation:

  The speed of an object, falling freely due to gravity, depends only on its height and not on its mass. Since the paths are frictionless and all the objects fall through the same height, therefore, their speeds on reaching the ground will be in the ratio of 1 : 1 : 1.
  Subject: 
  

  Physics
  Correct Answer: c
  

  Comment · 0 · 
• 5.

  In 1.0 s, a particle goes from point A to point B, moving in a semicircle of radius 1.0 m (see Figure). The magnitude of the average velocity

  
  a 
  

  3.14 m/s
  b 
  

  2.0 m/s
  c 
  

  1.0 m/s
  d 
  

  zero
  Difficulty Meter: 
  

  Intense (Only less than 20% of users made it right)
  Explanation:

  We know that,

  $\vec{\text{v}}=\frac{\text{Displacement of the particle}}{\text{Time taken}}=\frac{2.0m}{1s}=2.0m/s$
  Subject: 
  

  Physics
  Correct Answer: b
  

  Comment · 0 · 
• 6.

  Statement - 1: For an observer looking out through the window of a fast moving train, the nearby objects appear to move in the opposite direction to the train, while the distant objects appear stationary.

  Statement - 2 : If the observer and the object are moving at velocities $\overrightarrow{{\text{v}}_1}$ and  $\overrightarrow{{\text{v}}_2}$ respectively with reference to a laboratory frame, the velocity of the object with respect to the observer is $\overrightarrow{{\text{v}}_2}-\overrightarrow{{\text{v}}_1}.$
  a 
  

  Statement - 1 is true, Statement - 2 is true; Statement - 2 is correct explanation for Statement - 1.
  b 
  

  Statement -1 is true, Statement - 2 is true; Statement - 2 is not correct explanation for Statement - 1.
  c 
  

  Statement - 1 is true, Statement - 2 is false.
  d 
  

  Statement - 1 is false, Statement - 2 is true.
  Difficulty Meter: 
  

  Intense (Only less than 20% of users made it right)
  Explanation: 

  
  Subject: 
  

  Physics
  Correct Answer: b
  

  Comment · 0 · 
• 7.

  A man throws a ball vertically upward and it rises through 20 m and returns to his hands. The initial velocity (u) of the ball and for how much time (T) it remained in the air respectively are (g = 10m/s²)
  a 
  

  u = 10 m / s, T = 2s
  b 
  

  u = 10 m / s, T = 4s
  c 
  

  u = 20 m / s, T = 2s
  d 
  

  u = 20 m / s, T = 4s
  Difficulty Meter: 
  

  Intense (Only less than 20% of users made it right)
  Explanation:

  $u=\sqrt{2gh}=\sqrt{2\times 10\times 20}=20m/s$

  and $T=\frac{2u}{g}=\frac{2\times 20}{10}=4sec$
  Subject: 
  

  Physics
  Correct Answer: d
  

  Comment · 0 · 
• 8.

  From the top of a tower, a particle is thrown vertically downwards with a velocity of 10 ms-1. The ratio of the distances covered by it in the 3rd and 2nd seconds of the motion is (Take g = 10 ms^-3)
  a 
  

  5 : 7
  b 
  

  7 : 5
  c 
  

  3 : 6
  d 
  

  6 : 3
  Difficulty Meter: 
  

  Intense (Only less than 20% of users made it right)
  Explanation:

  $S_{3rd}=10+\frac{10}{2}(2\times 3-1)=35m$

  $S_{2nd}=10+\frac{10}{2}(2\times 2-1)=25m$

  $\Rightarrow$ $\frac{S_{3rd}}{S_{2nd}}=\frac{35}{25}=\frac{7}{5}=7:5$
  Subject: 
  

  Physics
  Correct Answer: b
  

  Comment · 0 · 
• 9.

  A car accelerates from rest at a constant rate a for sometime, after which it decelerates at a constant rate b and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is
  a 
  

  $\left(\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }\right)t$
  b 
  

  $\left(\frac{{\alpha }^2-{\beta }^2}{\alpha \beta }\right)t$
  c 
  

  $\frac{(\alpha +\beta )t}{\alpha \beta }$
  d 
  

  $\frac{\alpha \beta t}{\alpha +\beta }$
  Difficulty Meter: 
  

  Intense (Only less than 20% of users made it right)
  Explanation:

  Let the car accelerate at rate for time t 1 then maximum velocity attained, v = 0 + α t 1 = α t 1
  Now, the car decelerates at a rate for time (t – t₁) and finally comes to rest. Then,
  $0=v-\beta (t-t_1)\Rightarrow 0=\alpha t_1-\beta t+\beta t_1$

  $\Rightarrow t_1=\frac{\beta }{\alpha +\beta }t$

  $\therefore v=\frac{\alpha \beta }{\alpha +\beta }t$
  Subject: 
  

  Physics
  Correct Answer: d
  

  Comment · 0 · 
• 10.

  A self-propelled vehicle of mass m whose engine delivers constant power P has an acceleration $a=\frac{P}{mv}$(assume that there is no friction). In order to increase its velocity from v₁ to v₂ , the distance it has to travel will be
  a 
  

  
  b 
  

  $\frac{m}{3P}({\text{v}}_2^3-{\text{v}}_1^3)$
  c 
  

  $\frac{m}{3P}({\text{v}}_2^2-{\text{v}}_1^2)$
  d 
  

  $\frac{m}{3P}({\text{v}}_2-{\text{v}}_1)$
  Difficulty Meter: 
  

  Very Hard (About 20% - 40% of users made it right)
  Explanation:

  $\frac{d\text{v}}{dt}=\frac{P}{m\text{v}}$

  or $\text{v}\frac{d\text{v}}{ds}=\frac{P}{m\text{v}}$

  or ${\int }_0^{s}ds=\frac{m}{P}{\int }_{{\text{v}}_1}^{{\text{v}}_2}{\text{v}}^{2}d\text{v}$

  $\therefore s=\frac{m}{3P}({\text{v}}_2^3-{\text{v}}_1^3).$
  Subject: 
  

  Physics
  Correct Answer: b
  

  Comment · 0 · 
• 11.

  The graph of an object’s motion (along the x-axis) is shown in the figure. The instantaneous velocity of the object at points A and B are v_A and v_Brespectively. Then

  
  a 
  

  v_A = v_B = 0.5 m/s
  b 
  

  v_A = 0.5 m/s < v_B
  c 
  

  v_A = 0.5 m/s > v_B
  d 
  

  v_A = v_B = 2 m/s
  Difficulty Meter: 
  

  Intense (Only less than 20% of users made it right)
  Explanation:

  Instantaneous velocity  $\text{v}=\frac{\Delta x}{\Delta t}$

  From graph, ${\text{v}}_A=\frac{\Delta x_A}{\Delta t_A}=\frac{4m}{8s}=0.5m/s$

  and ${\text{v}}_B=\frac{\Delta x_B}{\Delta t_B}=\frac{8m}{16s}=0.5m/s$

  i.e., v_A = v_B = 0.5 m/s
  Subject: 
  

  Physics
  Correct Answer: a
  

  Comment · 0 · 
• 12.

  A body, thrown upwards with some velocity reaches the maximum height of 50 m. Another body with double the mass thrown up with double the initial velocity will reach a maximum height of
  a 
  

  100 m
  b 
  

  200 m
  c 
  

  300 m
  d 
  

  400 m
  Difficulty Meter: 
  

  Intense (Only less than 20% of users made it right)
  Explanation:

  The height attained by the body will not depend upon mass as deceleration in both the cases will be same.

  Applying, ${\text{v}}^2=u^2-2gh\Rightarrow 0=u^2-2gh$

  $h=\frac{u^2}{2g};h_1=50=\frac{u^2}{2g}$

  $h_2=\frac{(2u{)}^2}{2g}=\frac{4u^2}{2g}=4\times 50=200m$
  Subject: 
  

  Physics
  Correct Answer: b
  

  Comment · 0 · 
• 13.

  An express train is moving with a velocity v₁. Its driver finds another train is moving on the same track in the opposite direction with velocity v₂. To escape collision, driver applies a retardation a on the train. The minimum time for escaping collision will be :
  a 
  

  $t=\frac{{\text{v}}_1-{\text{v}}_2}{a}$
  b 
  

  $t=\frac{{\text{v}}_1^2-{\text{v}}_2^2}{2}$
  c 
  

  both (a) and (b)
  d 
  

  None of these
  Difficulty Meter: 
  

  Intense (Only less than 20% of users made it right)
  Explanation:

  v₁₂ = u₁₂ + a₁₂ t

  0 = (v₁ – v₂ ) – at

  $\therefore t=\left[\frac{{\text{v}}_1-{\text{v}}_2}{a}\right]$
  Subject: 
  

  Physics
  Correct Answer: a
  

  Comment · 0 · 
• 14.

  Statement 1 : Velocity–time graph for an object in uniform motion along a straight path is a straight line parallel to the time axis.
  Statement 2 : In uniform motion of an object velocity increases as the square of time elapsed.
  a 
  

  Statement -1 is false, Statement-2 is true
  b 
  

  Statement -1 is true, Statement-2 is true; Statement -2 is
  a correct explanation for Statement-1
  c 
  

  Statement -1 is true, Statement-2 is true; Statement -2
  is not a correct explanation for Statement-1
  d 
  

  Statement -1 is true, Statement-2 is false
  Difficulty Meter: 
  

  Intense (Only less than 20% of users made it right)
  Explanation:

  In uniform motion the object moves with uniform velocity,the magnitude of its velocity at different instane i.e., at t = 0, t =1, sec, t = 2sec ..... will always be constant. Thus velocity-time graph for an object in uniform motion along a straight path is a straight line parallel to time axis.
  Subject: 
  

  Physics
  Correct Answer: d
  

  Comment · 0 · 
• 15.

  A car, starting from rest, accelerates at the rate $f$  through a distance s, then continuous at constant speed for time t and then decelerates at the rate $\frac{f}{2}$ and come to rest. If the total distance traversed is 5 s, then :
  a 
  

  $s=\frac{1}{4}f{t}^2$
  b 
  

  $s=\frac{1}{2}f{t}^2$
  c 
  

  $s=\frac{1}{6}f{t}^2$
  d 
  

  $s=ft$
  Difficulty Meter: 
  

  Very Hard (About 20% - 40% of users made it right)
  Explanation:

  

  $s=\frac{1}{2}f{t}_1^2;\therefore t_1=\sqrt{\frac{2s}{f}}$

  ${\text{v}}_{max}=f{t}_1=f\sqrt{\frac{2s}{f}}=\sqrt{2fx}$

  Thus  $5s=\frac{1}{2}\left[\right(t+3t_1)+t]\times {\text{v}}_{max.}$

  or  $5s=\frac{1}{2}(2t+3t_1)\times {\text{v}}_{max.}$

  or  $5s=\frac{1}{2}\left(2t+3\sqrt{\frac{2s}{f}}\right)\times \sqrt{2fs}$

  $\therefore s=\frac{1}{2}f{t}^2$
  Subject: 
  

  Physics
  Correct Answer: b
  

  Comment