Tag: modelling collisions

Questions Related to modelling collisions

 Assertion (A) : A body of "$m _{1}$" collides another body of mass "$m _{2}$" at rest elastically. The fraction of energy transferred to the second body is$\dfrac{m _{1}}{m _1+m _{2}}$
Reason (R) : In an "inelastic collision" only linear momentum is conserved

modelling collisions collisions momentum work, energy and power physics

  1. Both Assertion (A) and Reason (R) are correct

    and R is the correct explanation

  2. Both Assertion (A) and Reason (R) are correct but

    the reason does not give the correct explanation

  3. A is true but R is false

  4. A is false but R is true

Show answer
Correct Option: D
Explanation:

In elastic collision no energy loss takes place that is total energy is transferred.
However, in inelastic collision, energy loss takes place and only linear momentum is conserved.

The graph between applied force and change in the length of wire within elastic limit is a ___________.

modelling collisions collisions momentum work, energy and power physics

  1. Straight line with positive slope

  2. Straight line with negative slope

  3. Curve with positive slope

  4. Curve with negative slope

Show answer
Correct Option: A
Explanation:

According to the Hooke's law,

$Y=\dfrac{Stress}{Strain}=\dfrac{F/A}{\Delta x/l}$
$\implies F=\dfrac{YA\Delta x}{l}$
$\implies F\propto \Delta x$
Hence correct answer is option A.

During inelastic collision between two bodies, which of the following quantities always remain conserved?

modelling collisions collisions momentum work, energy and power physics

  1. Total kinetic energy.

  2. Total mechanical energy.

  3. Total linear momentum.

  4. Speed of each body

Show answer
Correct Option: C
Explanation:

Since no external forces are acting on the colliding bodies during collision, thus total linear momentum is always conserved in all type of collisions but kinetic energy in not conserved in all collisions.

Kinetic energy is conserved in perfectly elastic collision only but some kinetic energy is lost in inelastic collisions. So, total kinetic energy is not conserved in inelastic collision.

In an inelastic collision, the kinetic energy after collision

modelling collisions collisions momentum work, energy and power physics

  1. is same as before collision

  2. is always less than that before collision

  3. is always greater than that before collision

  4. may be less or greater than that before collision

Show answer
Correct Option: B
Explanation:

In inelastic collisions , the kinetic energy is used for deforming the the bodies . In such collisions , the reformation of the bodies is partial. Therefore, the potential energy stored in the deformation is lost as heat and the kinetic energy of the system after collision is less than that of before collisions. However, in some cases it may be greater ( such collisions are called super elastic collisions).

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

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

Show answer
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

Show answer
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.