Tag: collisions in one dimension

Questions Related to collisions in one dimension

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) $

A ball with mass m and speed $V _0$ hit a wall and rebounds back with same speed.
Calculate the change in the object's kinetic energy.

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

  1. $-mv _0 ^2$

  2. $- \frac{1}{2}mv _0 ^2$

  3. Zero

  4. $ \frac{1}{2}mv _0 ^2$

  5. $mv _0 ^2$

Show answer
Correct Option: C
Explanation:

The speed of the ball remains the same $v _0$ before and after the collision with wall.

Thus the kinetic energy remains $\dfrac{1}{2}mv _0^2$.
Since kinetic energy is a scalar quantity, the change in it is zero.

The coefficient of restitution (e) for 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: D
Explanation:

$(O.R/e)=\dfrac{Relative:speed:after:collision}{Relative:speed:before:collision}$

For an elastic collision coefficient of restitution is 1.

 A body of mass $m$ moving at a constant velocity $v$ hits another body of the same mass moving at the same velocity but in the opposite direction and sticks to it. The common velocity after collision is

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

  1. $v$

  2. $0$

  3. $2v$

  4. $\dfrac{v}{2}$

Show answer
Correct Option: B
Explanation:

Since the collision is inelastic, applying momentum conservation for inelastic collisions,
$ mv + m(-v) = (m+m)V $
$ V = 0 $.

The co-efficient of restitution for a perfectly elastic collision is:

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

  1. $1$

  2. $0$

  3. lies in between $0$ and $1$

  4. infinity

Show answer
Correct Option: A
Explanation:

The coefficient of restitution is defined as the ratio of the relative velocity of separation to that of approach, in a situation of two objects colliding with each other. 

The relative velocity of approach is the difference between the individual velocities of the two bodies before the collision.
The relative velocity of separation is that after the collision.
In a perfectly elastic collision, the two relative velocities are exactly equal. Hence the coefficient becomes $= 1$

A lighter body moving with a velocity $v$ collides with a heavier body at rest. Then :

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

  1. the lighter body rebounces with twice the velocity of bigger body

  2. the lighter body retraces its path with the same velocity in magnitude

  3. the heavier body does not move practically

  4. both (2) and (3)

Show answer
Correct Option: D
Explanation:

In a collision system where $m _1$ moves with $u _1$ initially and $m _2$ is at rest with $m _2 >>> m _1$.
Let the final velocity of $m _1$ be $v _1$ and $m _2$ be $v _2$.
Assumption: Let the collision be elastic. Using linear momentum conservation and equation for coefficient of restitution,
$ m _1u _1 = m _2v _2 + m _1v _1 $
$ v _2 - v _1 = u _1 $


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

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

Using  $m _2 >>>  m _1,$  we get $ v _1 = -u _1 $ and $ v _2 = 0 $.  

Two identical bodies moving in opposite direction with same speed, collided with each other. If the collision is perfectly elastic then

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

  1. after the collision both comes to rest

  2. after the collision first comes to rest and second moves in the same direction with a speed 2v

  3. after collision they recoil with same speed

  4. all the above are possible

Show answer
Correct Option: C
Explanation:

$mv+m(-v)=m{ v } _{ 1 }+m{ v } _{ 2 }=0\Rightarrow { v } _{ 1 }+{ v } _{ 2 }=0\$
$ \dfrac { { v } _{ 1 }-{ v } _{ 2 } }{ v-(-v) } =1\Rightarrow { v } _{ 1 }-{ v } _{ 2 }=2v\ $

$thus,\quad { v } _{ 1 }=v\ { v } _{ 2 }=-v$

 A 6 kg mass travelling at $2.5 ms^{-1}$ collides head on with a stationary 4 kg mass. After the collision the 6 kg mass travels in its original direction with a speed of $1 ms^{-1}$. The coefficient of restitution is

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

  1. $1/4$

  2. $1/2$

  3. $3/4$

  4. $5/8$

Show answer
Correct Option: B
Explanation:

$e=\dfrac{v-1}{2.5}$

$=\dfrac{1.25}{2.5}=\dfrac{1}{2}$

A particle of mass m is attached to one end of a massless spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t=0$ with an initial velocity $u _0$. When the speed of the particle is $0.5 u _0$. It collides elastically with a rigid wall. After this collision.

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

  1. The speed of the particle when it returns to its equilibrium position is $u _0$

  2. The time at which the particle passes through the equilibrium position for the first time is $t=\pi\sqrt{\dfrac{m}{k}}$

  3. The time at which the maximum compression of the spring occurs is $t=\dfrac{4\pi}{3}\sqrt{\dfrac{m}{k}}$

  4. The time at which the particle passes through the equilibrium position for the second time is $t=\dfrac{5\pi}{3}\sqrt{\dfrac{m}{k}}$

Show answer
Correct Option: A,D
Explanation:

$\cfrac{1}{2}mv _o^2=\cfrac{1}{2}Kx^2+\cfrac{1}{2}\times m(0.25)v _o^2\quad equation (1)$

After elastic collision, black speed $=v _o$
So, when, it comes back to equilibrium point its speed is $u _o$.
Amplitude, $A=\cfrac {u _o}{\sqrt{K}}$
From equation $(1)$
$x=\cfrac{\sqrt {3}u _o}{2}\sqrt {\cfrac{m}{K}}$
$\therefore t _1=\cfrac {\pi}{3\omega}=\cfrac {\pi\sqrt{m}}{3\sqrt {K}}$
Time to reach the equilibrium position for the first time $=\cfrac{2\pi}{3} \sqrt {\cfrac{m}{K}}$
Second time, it will reach at time
$=\cfrac{2\pi}{3}\sqrt {\cfrac{m}{K}}+\cfrac {T}{2}$
$=\cfrac{2\pi}{3}\sqrt{\cfrac{m}{K}}+\cfrac {2\pi\sqrt {m}}{\sqrt {K}\pi 2}$
$=\cfrac {5\pi}{3}\sqrt {\cfrac {m}{K}}$

A thin uniform rod of mass $m$ and length $l$ is hinged at the lower end of a level floor and stands vertically. It is now allowed to fall, then its upper and will strike the floor with a velocity given by(A)$\sqrt { mgl }$(B) $\sqrt { 3gl }$(c)$\sqrt { 5gl }$ (D) $\sqrt { 2gl }$  Sol. 

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

  1. A

  2. B

  3. C

  4. D

Show answer
Correct Option: B