Tag: collisions

Questions Related to collisions

A ball P moving with a speed of $v \ ms^{-1}$ collides directly with another identical ball Q moving with a speed $10\ ms^{-1}$ in the opposite direction. P comes to rest after the collision. If the coefficient of restitution is 0.6, the value of $v$ is:

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

  1. $30\ ms^{-1}$

  2. $40\ ms^{-1}$

  3. $50\ ms^{-1}$

  4. $60\ ms^{-1}$

Show answer
Correct Option: B
Explanation:


As momentum is conserved, we can say,

$m(v-10)=mv _2$

$v _2=(v-10)$

$e=\dfrac{v _2-v _1}{u _1+u _2}=\dfrac{(v-10)-0}{(v+10)}$

$0.6=\dfrac{v-10}{v+10}$

$0.6v+6=v-10$

$0.4v=16$ 

$v=40\ ms^{-1}$

A ball is dropped from a $45\ m$ high tower while another is simultaneously thrown upward from the foot at $20\ m/s$, along the same vertical line. If the collision is perfectly elastic, first ball reaches ground after time-

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

  1. $2s$

  2. $3s$

  3. $4s$

  4. $5s$

Show answer
Correct Option: A

A body of mass $4m$ at rest explodes into three pieces. Two of the pieces each of mass $m$ move with a speed $v$ each in mutually perpendicular directions. The total kinetic energy released is:

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

  1. $\cfrac{1}{2}m{v}^{2}$

  2. $m{v}^{2}$

  3. $\cfrac{3}{2}m{v}^{2}$

  4. $\cfrac{5}{2}m{v}^{2}$

Show answer
Correct Option: A

A particle of mass m moving with velocity ${u} _{1}$ collides elastically with particle of same mass moving with velocity ${u} _{2}$ in the same direction. After collision their speeds are ${v} _{1}$ and ${v} _{2}$ respectively then-
(A) ${ u } _{ 1 }+{ v } _{ 1 }={ v } _{ 2 }+{ u } _{ 2 }$
(B)${ u } _{ 1 }-{ v } _{ 1 }={ v } _{ 2 }+{ u } _{ 2 }$

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

  1. Both the equations A and B are correct

  2. Both the equations A and B are incorrect

  3. Equation A is correct but not B

  4. Equation B is correct but not A

Show answer
Correct Option: C

A particle of mass $1\ kg$ moving with a velocity of $(4\hat {i}-3\hat {j})m/s$ collides with a fixed surface. After the collision velocity of the particle is $(4\hat {i}-3\hat {j})m/s$. Collision is

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

  1. Elastic

  2. Ineleastic

  3. Perfectly inelastic

  4. Data

Show answer
Correct Option: A

Two masses $m _{1}$ and $m _{2}$, approaches each other with equal speeds and collide elastically. After collision $m _{2}$ comes to rest. Then $m _{1}$/$m _{2}$ is

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

  1. $1$

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

  3. $\dfrac{1}{3}$

  4. $\dfrac{1}{4}$

Show answer
Correct Option: B

Two identical balls each of mass in are moving in opposite direction with a speed v. if they collide elastically maximum potentail energy stored in the ball is :

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

  1. 0

  2. $\dfrac { 1 }{ 2 } { mv }^{ 2 }$

  3. ${ mv }^{ 2 }$

  4. $2{ mv }^{ 2 }$

Show answer
Correct Option: A
Explanation:

Net momentum before collision will be $mv+(-mv)=0$, $negative$ because from $opposite $ direction.

so after the collision they will get stopped to make the net momentum again $zero$ and whole energy will be get stored
 as $PE$. ($inelastic $ $ collision$)

There is one other possibility too that is they $exchange$ their velocities so that again the net momentum will become
 $zero.$ $elastic$ $ collision$ .
In elastic collision there is no loss in $KE$  so $no$ storage of $PE.$


Two particles moving initially in the same direction undergo a one dimensional,elastic collision. Their relative velocities before and after the collision are $\overrightarrow { { v } _{ 1 } } $ and $\overrightarrow { { v } _{ 2 } } $. Then:

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

  1. $\left| \overrightarrow { { v } _{ 1 } } \right| = \left| \overrightarrow { { v } _{ 2 } } \right| $

  2. $\overrightarrow { { v } _{ 1 } } = - \overrightarrow { { v } _{ 2 } }$ only if the two are of equal mass.

  3. $\overrightarrow { { v } _{ 1 } } = -\overrightarrow { { v } _{ 2 } } = {\left| \overrightarrow { { v } _{ 1 } } \right|}^{2}$

  4. $\left| \overrightarrow { { v } _{ 1 }} . \overrightarrow { { v } _{ 2 }} \right| = -  {\left| \overrightarrow { { v } _{ 1 } } \right|}^{2}$

Show answer
Correct Option: D

The coefficient of restitution of a perfectly elastic collision is :

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

  1. $1$

  2. $0$

  3. $\infty$

  4. $-1$

Show answer
Correct Option: A
Explanation:

The Coefficent of Restitution is a measure of the "bounciness" of a collision between two objects: how much of the kinetic energy remains for the objects to rebound from one another vs. how much is lost as heat, or work done deforming the objects.

The coefficient, e is defined as the ratio of relative speeds after and before an impact, taken along the line of the impact:
$e=\dfrac { Speed\quad of\quad separation }{ Speed\quad of\quad approach } $

(i)   For perfectly elastic collision $e = 1$
(ii)  For perfectly inelastic collision $e = 0$
(iii) For other collision $0 \lt e \lt 1$

A ball moving with a velocity v strikes a wall moving toward the ball with a velocity u. An elastic impact lasts for t sec. Then the mean elastic force acting on the ball is 

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

  1. $\displaystyle \frac { 2mv }{ t } $

  2. $\displaystyle \frac { 2m\left( \upsilon +u \right) }{ t } $

  3. $\displaystyle \frac { 2m\left( \upsilon +2u \right) }{ t } $

  4. $\displaystyle \frac { m\left( 2\upsilon +u \right) }{ t } $

Show answer
Correct Option: B
Explanation:

Relative speed of the ball $\displaystyle =\left( \upsilon +u \right) $
Speed after rebouncing $\displaystyle =-\left( \upsilon +u \right) $
Now, $\displaystyle F=m\frac { \Delta \upsilon  }{ \Delta t } =\frac { m }{ t } \left[ \left( \upsilon +u \right)  \right] -[-\left[ \left( \upsilon +u \right)  \right]] $
$\displaystyle =\frac { 2m }{ t } \left( \upsilon +u \right) $