Tag: work, energy and power

Questions Related to work, energy and power

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

modelling collisions collisions momentum work, energy and power physics

  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

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Correct Option: C
Explanation:

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

Choose the false statement

modelling collisions collisions momentum work, energy and power physics

  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.

Show answer
Correct Option: B
Explanation:

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

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 :

modelling collisions collisions momentum work, energy and power physics

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

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

  3. $\sqrt {2gH} $

  4. None of these

Show answer
Correct Option: B
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.

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:

modelling collisions collisions momentum work, energy and power physics

  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. $\vec{r _1}\times \vec{v _1}=\vec{r _2}\times \vec{v _2}$

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Correct Option: A

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 

modelling collisions collisions momentum work, energy and power physics

  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

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Correct Option: A

In an inelastic collision-

modelling collisions collisions momentum work, energy and power physics

  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.

Show answer
Correct Option: D
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.

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.

modelling collisions collisions momentum work, energy and power physics

  1. $v$

  2. $\sqrt { 3v } $

  3. $\frac { 2 }{ \sqrt { 3 } } v$

  4. $\frac { v }{ \sqrt { 3 } } $

Show answer
Correct Option: C

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 ?

modelling collisions collisions momentum work, energy and power physics

  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 $

Show answer
Correct Option: B
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:
$\dfrac {1}{2}m _{1}u _{1}^{2}+\dfrac {1}{2}m _{2}u _{2}^{2}=\dfrac {1}{2}m _{1}v _{1}^{2}+\dfrac {1}{2}m _{2}v _{2}^{2}$
$2\times 16+0=2\times 1+m _{2}v _{2}^{2}$
$m _{2}v _{2}^{2}=30$
$\dfrac {m _{2}v _{2}^{2}}{m _{2}v _{2}}=\dfrac {30}{6}$
$v _{2}=5\ m/s$
$m _{2}=1.2\ kg$


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 :

modelling collisions collisions momentum work, energy and power physics

  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) $

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Correct Option: B

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

modelling collisions collisions momentum work, energy and power physics

  1. $mnu$

  2. $2\ mnu$

  3. $4\ mnu$

  4. $mnu/2$

Show answer
Correct Option: B
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 $F=\dfrac{Momentum}{time}=\dfrac{2mnu}{1sec}=2mnu$