Question

A ball of mass $ m $ moving with velocity $ {{v}_{o}} $ experiences a head-on elastic collision with one of the spheres of a stationary rigid dumbbell as shown in figure. The mass of each sphere equals $ \dfrac{m}{2} $ , and the distance between them is $ l $ .Disregarding the size of the spheres, find the proper angular momentum $ \overrightarrow{M} $ of the dumbbell after the collision, i.e. the angular momentum in the reference frame moving translationally and fixed to the dumbbell's centre of inertia is given as $ \overrightarrow{M}=\dfrac{m{{v}_{o}}l}{x} $ . Find $ x $ .

Answer 1

Hint: For a head-on elastic collision, momentum and kinetic energy remains conserved. We will apply the expression for conservation of linear momentum of the system during elastic collision to calculate the required value of $ x $ . The values of velocity and momentum can be calculated using the equation of coefficient of restitution during collision.

Formula used:
Conservation of linear momentum:
$ {{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}} $
Coefficient of restitution, $ e=\dfrac{\left| {{v}_{1f}}-{{v}_{2f}} \right|}{\left| {{v}_{1i}}-{{v}_{2i}} \right|} $

Complete step-by-step answer:
In a head on elastic collision, linear momentum of the system remains conserved.
Law of conservation of momentum states that the total momentum of the system remains constant (or conserved), if the external force acting on the system is zero.
Conservation of linear momentum:
 $ {{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}} $
Where,
 $ {{m}_{1}} $ is the mass of first particle
 $ {{m}_{2}} $ is the mass of second particle
 $ {{u}_{1}} $ is the initial velocity of first particle
 $ {{u}_{2}} $ is the initial velocity of second particle
 $ {{v}_{1}} $ is the final velocity of first particle
 $ {{v}_{2}} $ is the final velocity of second particle
We are given that the collision is head-on and perfectly elastic.
From the law of conservation of linear momentum along the direction of incident ball, system consists of colliding ball and sphere, we have
 $ m{{v}_{o}}=mv'+\dfrac{m{{v}_{1}}}{2} $
Or,
 $ {{v}_{o}}=v'+\dfrac{{{v}_{1}}}{2} $
(Let’s say it equation 1)
Where,
 $ v' $ is the velocity of ball
 $ {{v}_{1}} $ is the velocity of sphere 1 after collision
As given the collision is perfectly elastic, we will use the equation of coefficient of restitution,
Coefficient of restitution is expressed as the ratio of the final relative velocity of two bodies to the initial relative velocity of the objects.
 $ e=\dfrac{\left| {{v}_{1}}-{{v}_{2}} \right|}{\left| {{u}_{1}}-{{u}_{2}} \right|} $
Where,
 $ {{u}_{1}} $ is the initial velocity of first particle
 $ {{u}_{2}} $ is the initial velocity of second particle
 $ {{v}_{1}} $ is the final velocity of first particle
 $ {{v}_{2}} $ is the final velocity of second particle
For a perfectly elastic collision $ e=1 $
Putting the values, we get,
 $ 1=\dfrac{v'-{{v}_{1}}}{0-{{v}_{o}}} $
Where,
 $ v' $ is the velocity of ball
 $ {{v}_{1}} $ is the velocity of sphere 1 after collision
Or,
 $ v'-{{v}_{1}}=-{{v}_{o}} $
(Let’s say it equation 2)
Solving the above equation 1 and 2,
 $ v'-{{v}_{1}}=-{{v}_{o}} $
 $ {{v}_{o}}=v'+\dfrac{{{v}_{1}}}{2} $
We get,
 $ {{v}_{1}}=\dfrac{4{{v}_{o}}}{3} $ , directed towards right direction
After the collision, the dumbbell will rotate and acquire some angular momentum associated with its rotation. The dumbbell will rotate in such a way that the magnitude of angular momentum of two spheres will be the same but the direction will be opposite.
In the centre of mass frame of sphere 1 and sphere 2,
 $ \overrightarrow{{{p}_{1}}}=-{{\overrightarrow{{{p}_{2}}}}_{{}}} $
Where,
 $ \overrightarrow{{{p}_{1}}} $ is the linear momentum of sphere 1of dumbbell
 $ \overrightarrow{{{p}_{2}}} $ is the linear momentum of sphere 2 of dumbbell
 $ \left| \overrightarrow{{{p}_{1}}} \right|=\left| \overrightarrow{{{p}_{2}}} \right|=\mu \left| \overrightarrow{{{v}_{1}}}-\overrightarrow{{{v}_{2}}} \right| $
Where,
 $ {{v}_{1}} $ is the velocity of sphere 1 of dumbbell after collision
 $ {{v}_{2}} $ is the velocity of sphere 2 of dumbbell after collision
 $ \mu $ is the mass of centre of mass of dumbbell system
Also,\[\overrightarrow{{{r}_{1}}}=-\overrightarrow{{{r}_{2}}}\]
Where,
 $ {{r}_{1}} $ is the distance of sphere 1 of dumbbell from the centre of mass
 $ {{r}_{2}} $ is the distance of sphere 2 of dumbbell from the centre of mass
Therefore,
\[\widetilde{\overrightarrow{M}}=2\left[ \overrightarrow{{{r}_{1}}}\times \overrightarrow{{{p}_{1}}} \right]\]
Where,
 $ \overrightarrow{M} $ is the angular momentum of the dumbbell in the reference frame, that is, the centre of mass frame of dumbbell.
Also,
 $ \overrightarrow{{{r}_{1}}}\bot \widetilde{\overrightarrow{{{p}_{1}}}} $
We get,
 $ \widetilde{\overrightarrow{M}}=2\left[ \dfrac{l}{2}\text{ }\dfrac{\dfrac{m}{2}}{2}\text{ }\dfrac{4{{v}_{o}}}{3}\widehat{n} \right] $
Where $ \widehat{n} $ is the unit vector in the sense of $ \overrightarrow{{{r}_{1}}}\bot \widetilde{\overrightarrow{{{p}_{1}}}} $
Therefore,
 $ \widetilde{M}=\dfrac{m{{v}_{o}}l}{3} $
Angular Momentum of the dumbbell after the collision is $ \widetilde{M}=\dfrac{m{{v}_{o}}l}{3} $
Comparing $ \dfrac{m{{v}_{o}}l}{3} $ with $ \dfrac{m{{v}_{o}}l}{x} $
We get $ x=3 $

Hence, the value of $ x $ is 3.

Additional information:
Collision, also known as impact, is the sudden and forceful coming together in direct contact of two bodies. Two types of collisions can be Elastic collision and Non-elastic collision.
In elastic collision, total kinetic energy of two bodies, that is our system, remains conserved. In an ideal elastic collision, known as a perfectly elastic collision, there is no net transformation of kinetic energy into other forms of energy such as heat, noise or potential energy. Also, momentum is conserved in an elastic collision. Momentum of the whole system before collision is equal to the momentum of the system after collision. In non-elastic collisions, momentum of the system remains conserved. But there is a loss in kinetic energy of the system.
A head-on, or direct, collision means that the point of impact is on the straight line connecting the centre of gravity of each of the body. A head-on collision is the one where the front ends two bodies hit each other, as opposed to a side collision or rear end collision.
Coefficient of restitution is expressed as the ratio of the final relative velocity of two bodies to the initial relative velocity of the objects.
 $ e=\dfrac{\left| {{v}_{1f}}-{{v}_{2f}} \right|}{\left| {{v}_{1i}}-{{v}_{2i}} \right|} $
Value of the coefficient of restitution varies from 0 to 1.
For a perfectly elastic collision $ e=1 $
For a perfectly inelastic collision $ e=0 $

Note: Students should know the concept of head-on and elastic collision, for solving the above question. Also remember, in a perfectly elastic collision the momentum and the kinetic energy of the system remains conserved. Value of the coefficient of restitution in a perfectly elastic collision is one.