Tag: momentum

Quizzes (4) Questions (63)

Questions Related to momentum

Multiple choice modelling collisions collisions momentum work, energy and power physics

A ball hits the floor and rebounds after an inelastic collision. In this case

  1. the momentum of the ball just after the collision is same as that just before the collision

  2. The mechanical energy of the ball remains the same in the collision

  3. the total momentum of the ball and the earth is conserved

  4. the total energy of the ball and the earth remains the same
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

1) K.E will be less since collision is inelastic
2) No impulsive external force on system (earth + ball)

Multiple choice modelling collisions collisions momentum work, energy and power physics

Choose the false statement

  1. In a perfect elastic collision the relative velocity of approach is equal to the relative velocity of separation

  2. In an inelastic collision the relative velocity of approach is less than the relative velocity of separation

  3. In an inelastic collision the relative velocity of separation is less than the relative velocity of approach

  4. In perfect inellastic collision relative velocity of separation is zero.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In an inelastic collision the relative velocity of approach is more than the relative velocity of separation. Hence B is wrong

Multiple choice modelling collisions collisions momentum work, energy and power physics

Two pendulum bobs of mass $m$ and  $2\ m$ collide elastically at the lowest point in their motion. If both the balls are released from height $H$ above the lowest point. The velocity of the bob of mass $m$ just after collision is :

  1. $\sqrt {\dfrac{{2gH}}{3}} $
  2. $\dfrac{5}{3}\sqrt {2gH} $
  3. $\sqrt {2gH} $

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Lets consider first pendulum bob of mass $m _1$ and second  pendulum bob of mass $m _2$ collide elastically at the lower point in their motion. If both are released  from the height $H$ above the lower point.


At height $H$, Kinetic energy is equal to potential energy.


$ \dfrac{1}{2}mu^2=mgH$

$u=\sqrt{2gH}$

Given ,

$m _1=m$

$m _2=2m$

$u _1=-\sqrt{2gH}$

$u _2=\sqrt{2gH}$

In elastic collision,

The velocity of first $v _1$ after collision at the lower point,

$v _1=(\dfrac{m _1-m _2}{m _1+m _2})u _1+(\dfrac{2m _2}{m _1+m _2})u _2$

By putting the given value of $m _1$, $m _2$, $u _1$, $u _2$ in the above equation,

$v _1=(\dfrac{m-2m}{m+2m})(-\sqrt{2gH})+(\dfrac{4m}{m+2m})\sqrt{2gH}$

$v _1=\dfrac{5}{3}\sqrt{2gH}$

Thus, the correct option is B.

Multiple choice modelling collisions collisions momentum work, energy and power physics

Two particles A and B, move with constant velocities $\vec{v _1}$ and $\vec{v _2}$. At the initial moment their position vectors are $\vec{r _1}$ and $\vec{r _2}$ respectively. The condition for particle A and B's collision is:

  1. $\vec{r _1}-\vec{r _2}=\vec{v _1}-\vec{v _2}$
  2. $\dfrac{\vec{r _1}-\vec{r _2}}{|\vec{r _1}-\vec{r _2}|}=\dfrac{\vec{v _2}-\vec{v _1}}{|\vec{v _2}-\vec{v _1}|}$
  3. $\vec{r _1}\cdot \vec{v _1}=\vec{r _2}\cdot \vec{v _2}$
  4. $\vec{r _1}\times \vec{v _1}=\vec{r _2}\times \vec{v _2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Two particles will collide if their relative position vector is parallel (or antiparallel) to their relative velocity vector. This condition is expressed mathematically by setting the unit vector of relative position equal to the negative of the unit vector of relative velocity, which simplifies to the ratio of position difference vectors matching the direction of velocity difference.

Multiple choice modelling collisions collisions momentum work, energy and power physics

A body of mass 2kg is projected upward from the surface of the ground at t$=$0 with a velocity of 20 m/s. One second later a body B, also of mass 2 kg, is dropped from a height of 20 m. If they collide elastically, then velocities just after collision are 

  1. V$ _A = $ -5 m/s downward , V$ _B = $ 5 m/s upward
  2. V$ _A = $ 10 m/s downward , V$ _B = $ 5 m/s upward
  3. V$ _A = $ 10 m/s upward , V$ _B = $ 10 m/s downward

  4. both move downward with velocity 5 m/s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At t=1s, A is at y = 20(1) - 0.5(10)(1)^2 = 15m, velocity vA = 20 - 10(1) = 10 m/s. B is at y = 20 - 0.5(10)(0)^2 = 20m, velocity vB = 0 m/s. They collide at some point. Since masses are equal and collision is elastic, they exchange velocities. The calculation of the exact collision point and post-collision velocities confirms option A.

Multiple choice modelling collisions collisions momentum work, energy and power physics

In an inelastic collision-

  1. Momentum of the system is always conserved.

  2. Velocity of separation is less than the velocity of approach.

  3. The coefficient of restitution can be zero.

  4. All of the above.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In inelastic collision momentum of the system is always conserved if $F _{ext}=0$.
Velocity of separation is less than the velocity of approach since co-efficient of restitution e < 1
$e=0$ for a perfectly inelastic collision since the colliding particles stick together after collision.
Hence, option (D) is correct.

Multiple choice modelling collisions collisions momentum work, energy and power physics

A mass $m$ moves with velocity $v$ and collides inelastically with another identical mass. After collision, the 1st mass moves with velocity $\frac { v }{ \sqrt { 3 }  } $ in a direction perpendicualr to the initial direction of motion. find the speed of the second mass after collision.

  1. $v$
  2. $\sqrt { 3v } $
  3. $\frac { 2 }{ \sqrt { 3 } } v$
  4. $\frac { v }{ \sqrt { 3 } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Conservation of momentum: m*v*i = m*v1*j + m*v2*cos(theta)*i + m*v2*sin(theta)*j. Since v1 = v/sqrt(3) in the y-direction, the x-component of momentum is conserved: m*v = m*v2*cos(theta), and y-component: 0 = m*v1 + m*v2*sin(theta). Solving these gives v2 = 2*v/sqrt(3).

Multiple choice modelling collisions collisions momentum work, energy and power physics

A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed v. What is the mass of other body and the speed of the center of mass of two bodies ?

  1. $ 1.0 kg and \frac {2}{3}v $
  2. $ 1.2 kg and \frac{5}{8} $
  3. $ 1.4 kg and \frac {10}{17} v $
  4. $ 1.5 kg and \frac {4}{7} v $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$m _{1}=2\ kg$
$u _{1}=4\ m/s$
$v _{1}=\dfrac {u _{1}}{4}=\dfrac {4}{4}=1\ m/s$
$u _{2}=0$
since nothing is given so we would take it as elastic collision. 
hence conservation of momentum is given by:
$m _{1}u _{1}+m _{2}u _{2}=m _{1}v _{1}+m _{2}v _{2}$
$2\times 4+0=2\times 1+m _{2}v _{2}$
$m _{2}v _{2}=6$
according to the conservation kinetic energy:
$\dfrac {1}{2}m _{1}u _{1}^{2}+\dfrac {1}{2}m _{2}u _{2}^{2}=\dfrac {1}{2}m _{1}v _{1}^{2}+\dfrac {1}{2}m _{2}v _{2}^{2}$
$2\times 16+0=2\times 1+m _{2}v _{2}^{2}$
$m _{2}v _{2}^{2}=30$
$\dfrac {m _{2}v _{2}^{2}}{m _{2}v _{2}}=\dfrac {30}{6}$
$v _{2}=5\ m/s$
$m _{2}=1.2\ kg$


Multiple choice modelling collisions collisions momentum work, energy and power physics

Two objects of mass 3 kg and 2 kg move along x and y ais with $ 4 m/s^2 $ and $ 3 m/s ^2 $ respectively on a horizontal smooth table.after collision the bodies stick together.then :

  1. Heat generated in the process is 15 joules

  2. Heat generated in the process is 18 joules
  3. direction of motion x-axis after collision is $ 60^0 $
  4. direction of x-axis after collision is $ tan{-1} \left( \frac { 1 }{ 3 } \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice modelling collisions collisions momentum work, energy and power physics

$n$ balls each of mass $m$ impinge elastically each second on a surface with velocity $u$. The average force experienced by the surface will be

  1. $mnu$
  2. $2\ mnu$
  3. $4\ mnu$
  4. $mnu/2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
  • Elastically means the balls get bounced with $same$ speed i.e $u$
  • so the change in momentum will be $P _2-P _1=-mu-mu=-2mu$
  • so the momentum tranferred to the surface will be $-(-2mu)=2mu$
  • thus the total momentum transferred by $n$ balls will be $2mnu$
  • force $F=\dfrac{Momentum}{time}=\dfrac{2mnu}{1sec}=2mnu$