Tag: collisions

Quizzes (0) Questions (169)

Questions Related to collisions

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

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.

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

  3. Zero
  4. $ \frac{1}{2}mv _0 ^2$
  5. $mv _0 ^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

 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

  1. $v$
  2. $0$
  3. $2v$
  4. $\dfrac{v}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

  1. $1$
  2. $0$
  3. lies in between $0$ and $1$

  4. infinity
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $= 1$

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

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

  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)
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $.  

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

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

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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\ $

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

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

 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

  1. $1/4$
  2. $1/2$
  3. $3/4$
  4. $5/8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

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.

  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 $t=\dfrac{5\pi}{3}\sqrt{\dfrac{m}{k}}$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
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
$=\cfrac{2\pi}{3}\sqrt {\cfrac{m}{K}}+\cfrac {T}{2}$
$=\cfrac{2\pi}{3}\sqrt{\cfrac{m}{K}}+\cfrac {2\pi\sqrt {m}}{\sqrt {K}\pi 2}$
$=\cfrac {5\pi}{3}\sqrt {\cfrac {m}{K}}$

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

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. 

  1. A

  2. B

  3. C

  4. D
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using conservation of energy, the potential energy of the rod (mg*l/2) is converted into rotational kinetic energy (1/2 * I * omega^2). For a rod hinged at one end, I = ml^2/3. Thus, mg*l/2 = (1/2)*(ml^2/3)*omega^2. Solving for omega gives omega = sqrt(3g/l). The velocity of the tip is v = omega*l = sqrt(3gl).

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

Two balls of equal mass undergo head on collision while each was moving with speed $6\ m/s$. If the coefficient of restitution is $\dfrac{1}{3}$, the speed of each ball after impact will be

  1. $18\ m/s$
  2. $2\ m/s$
  3. $6\ m/s$
  4. $4\ m/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let speed of balls are $v _1$ and $v _2$.

There is no external force acting, momentum will be conserved. 
$m _1u _1+ m _2u _2= m _1v _1+m _2v _2$
$\Rightarrow m\times 6-m\times 6=mv _1+ mv _2$
$\Rightarrow v _1=-v _2$
Coefficient , $e=-\dfrac{v _1-v _2}{u _1-u _2}$  $\Rightarrow \dfrac{1}{3}= -\dfrac{v _1-v _2}{6+6}$  $\Rightarrow v _1=-2 m/s$