Tag: collisions

Questions Related to collisions

A ball of mass $m _{1}$ is moving with velocity $v$. It collides head on elastically with a stationary ball of mass $m _{2}$. The velocity of ball becomes $\dfrac{v}{3}$ after collision, then the value of the ratio $\dfrac{m _{2}}{m _{1}}$ is:

collisions in one dimension collisions work, energy and power mechanics physics

  1. $1$

  2. $2$

  3. $3$

  4. $4$

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Correct Option: B

A ball of mass m moving with velocity v collides elastically with another ball of identical mass coming from the opposite direction with velocity 2v. Their velocities after collision are :

collisions in one dimension collisions work, energy and power mechanics physics

  1. $-v,:2v$

  2. $-2v,:v$

  3. $v,:-2v$

  4. $2v,:-v$

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Correct Option: B
Explanation:

When bodies of same mass collide head on elastically then after collision they exchange their velocities.

A sphere $'P'$ of mass $'m'$ moving with velocity $'u'$ collides head-on with another sphere $'Q'$ of mass $'m'$ which is at rest. The ratio of final velocity of $'Q'$ to initial velocity of $'P'$ is
($e =$ coefficient of restitution)

collisions in one dimension collisions work, energy and power mechanics physics

  1. $\dfrac{e-1}{2}$

  2. ${\left[\dfrac{e+1}{2}\right]}^{{1}/{2}}$

  3. $\dfrac{e+1}{2}$

  4. ${\left[\dfrac{e+1}{2}\right]}^{2}$

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Correct Option: C
Explanation:

Here,  ${m} _{1} = {m} _{2} = m$,     ${u} _{1} = u$,    ${u} _{2} = 0$
Let ${v} _{1},   {v} _{2}$ be their velocities after collision.
According to principle of conservation of linear momentum
$mu + 0 = m\left({v} _{1}+{v} _{2}\right)$
or   ${v} _{1} + {v} _{2} = v$       ....(i)
By definition,   $e = \dfrac{{v} _{2} - {v} _{1}}{u-0}$
or   ${v} _{2} - {v} _{1} = eu$      .....(ii)
Adding equations (i) and (ii), we get
${v} _{2} = \dfrac{u\left(1+e\right)}{2}    \Rightarrow    \dfrac{{v} _{2}}{u} = \dfrac{1+e}{2}$

If two balls each of mass 0.06 kg moving in opposite directions with speed of $4\, m\, s^{-4}$ collide and rebound with same speed, then the impulse imparted to each ball due to other is:

collisions in one dimension collisions work, energy and power mechanics physics

  1. $0.48\, kg\, m\,s^{-1}$

  2. $0.53\, kg\, m\,s^{-1}$

  3. $0.8\, kg\, m\,s^{-1}$

  4. $0.92\, kg\, m\,s^{-1}$

Show answer
Correct Option: A
Explanation:

Mass = $0.06kg$


Velocity= $4 m/s$

Rebound velocity = $-4 m/s$

Impulse = change in momentum 

Impulse = $m (u) – m (v)$

Impulse = $0.06 \times 4 – (0.06 \times -4) $

Impulse = $0.24 – (-0.24)$

Impulse = $0.24+0.24$


Impulse = $0.48kgm/s$

 A ball of mass '$M$' moving with a velocity $\overrightarrow{V}$ collides head on elastically with another body of the same mass    '$M$' moving with a velocity $-\overrightarrow{V}$ in the opposite direction. After the collision :

collisions in one dimension collisions work, energy and power mechanics physics

  1. The velocities are exchanged by the two balls

  2. Both the balls come to rest

  3. Both of them move at right angles to the original line of motion

  4. One ball comes to rest and the other ball travels back with a velocity $2v$

Show answer
Correct Option: A
Explanation:

Since the collision is elastic, we know that when two bodies of same mass have an elastic collision their velocities get interchanged.
$m _1u _1+m _2u _2=m _1v _1+m _2v _2$


But $m _1=m _2=M,   u _1=V,   u _2=-V$

$\Rightarrow MV-MV=Mv _1+Mv _2$

$\Rightarrow v _1+v _2=0$.............(1)

The coefficient of restitution $(e)$ is given by,

$e=-\dfrac{v _1-v _2}{u _1-u _2}$


But for elastic collision,  $e=1$

$-\dfrac{v _1-v _2}{V+V}=1$

$\Rightarrow v _2-v _1=2V$..............(2)

From (1) and (2), we get,
$v _2=V$
$v _1=-V$

 A heavy steel ball of mass greater than 1 kg moving with a speed of 2m/ s collides head on with a stationary ping pong ball of mass less than 0.1 g. The collision is elastic. After the collision the ping pong ball moves approximately with a speed

collisions in one dimension collisions work, energy and power mechanics physics

  1. $ 2 m / s $

  2. $4 m/ s$

  3. $2\times10^{4}m / s$

  4. $2\times10^{3}m / s$

Show answer
Correct Option: B
Explanation:

Since the body is much heavy these won't be much change in velocity
& $e = 1$


i.e., $\dfrac{v-2}{0-2} = -1$

$\Rightarrow v = 4 m/s$

Consider the following statements A and B. Identify the correct choice in the given answer

 A : In a one - dimensional perfectly elastic collision between two moving bodies of equal masses, the  bodies merely exchange their velocities after collision

 B : If a lighter body at rest suffers perfectly elastic collision with a very heavy body moving with a certain velocity, after collision both travel with same velocity

collisions in one dimension collisions work, energy and power mechanics physics

  1. A and B are correct

  2. Both A and B are wrong

  3. A is correct B is wrong

  4. A is wrong B is correct

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Correct Option: C
Explanation:

A : Only possible case.
B : Since masses are not same velocities can't be same.

 Consider the following statements A and B and identify the correct answer:
$A :$ In an elastic collision, if a body suffers a head on collision with another of same mass at rest, the first body comes to rest while the other starts moving with the velocity of the first one.
$B :$ Two bodies of equal masses suffering a head-on elastic collision merely exchanges their velocities.

collisions in one dimension collisions work, energy and power mechanics physics

  1. A and B are true

  2. A and B are false

  3. A is true but B is false

  4. A is false but B is true

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Correct Option: A
Explanation:

For an elastic collision $($ using momentum conservation and coefficient of restitiution $)$,


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

$ v _2 = \dfrac{2m _1}{m _1+m _2}u$.

If $m _1=m _2,$ then$, v _1 = 0 ,v _2 = u$.
In general, the velocities of the bodies just get interchanged in an elastic collision if their masses are equal.

In one - dimensional head on collision, the relative velocity of approach before collision is equal to :

collisions in one dimension collisions work, energy and power mechanics physics

  1. relative velocity of separation after collision

  2. $e$ times relative velocity of separation after collision

  3. $1/e$ times relative velocity of separation after collision

  4. sum of the velocities after collision

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Correct Option: C
Explanation:

Coefficient of restitution is given by,

$e = \dfrac{\text{velocity of separation along line of impact}}{\text{Velocity of approach line of impact}}$

$\therefore e = \dfrac{v _2-v _1}{u _1- u _2}$

$\therefore u-1 - u _2 = \dfrac{1}{e} (v _2-v _1)$

$\therefore$ option C is correct.

A body of mass 'm' moving with certain velocity collides with another identical body at rest. If the collision is perfectly elastic and after the collision, if both the bodies move, they can move

collisions in one dimension collisions work, energy and power mechanics physics

  1. in the same direction

  2. in opposite direction

  3. in perpendicular direction

  4. making $45^o$ to each other

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Correct Option: C
Explanation:

Make your co-ordinate axes such that 1 is along the line joining the centres of the 2 just at time on collision and one, tangential. Now consider the velocity of the coming ball in 2 components of the coordinate axes we just assumed, say vn and vt(v-normal and v-tangential). Now during collision, v-tangential can't change as there is no impact along this direction. And collision along normal is just like collision in 1-D. The equations are exactly same. Hence the 1st particle will have 0 velocity along normal and second will move along normal. Hence, they'll move perpendicular to each other.