Question

Three balls A, B and C of masses 2kg, 4kg and 8kg, respectively, move along the same straight line and in the same direction, with velocities 4m/s, 1m/s and ¾ m/s. if A collides with B and subsequently B collides with C, find the velocity of ball A and B after collision, taking the coefficient of restitution as unity.
A) $V_A=3,V_B={\displaystyle \frac{9}{4}}$
B) $V_A=0,V_B=3$
C) $V_A=3,V_B=0$
D) $V_A=0,V_B=0$

Answer 1

Hint:The above question can be solved on considering the elastic collisions between the bodies. That is there is the conservation of linear momentum i.e. Momentum before collision is equal to momentum after collision.

Formula used:
1) $e={\displaystyle \frac{\left({\text{v}}_{\text{2}}\text{ - }{\text{v}}_{\text{1}}\right)}{\left(u_1-u_2\right)}}$
2) $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
Where,
e=coefficient of restitution
$m_1$= Mass of 1st body
$m_2$ = Mass of 2nd body
$u_1$ =Initial velocity of 1st body
$u_2$ = Initial velocity of the second body
$$v_1$$= Final velocity of the first body
$$v_2$$= Final velocity of the second body

Complete step-by-step answer:
We will start by considering the initial velocities of the balls. Let assume that ${\text{u}}_{\text{1}}\text{ and }{\text{u}}_{\text{2}}$ be the velocities of the bodies respectively before collision and $v_{1}and{v}_2$ are their final velocities after collision which are acting along same line in a same direction.
We have given that the coefficient of restitution is unity. The coefficient of restitution (e) is defined as the ratio of the relative velocity of separation $\left(v_2-v_1\right)$ after the collision to the relative velocity of approach $\left(u_1-u_2\right)$ before collision.
Thus, we can write the coefficient of restitution (e) for the balls as,
$e={\displaystyle \frac{\left(v_2-v_1\right)}{\left(u_1-u_2\right)}}$
We are provided with e=1 then the equation becomes
$1={\displaystyle \frac{\left(v_2-v_1\right)}{\left(u_1-u_2\right)}}$
On rearranging the equation, we get,
$\left(u_1-u_2\right)=\left(v_2-v_1\right)$
Now, Let $u_1$ and $u_2$ be the initial velocities of block A and B before collision and $v_1$ and $v_2$ be the final velocities after collision.
We are provided with the masses and velocities of block which are,
$$m_1=2kg,m_2=4kg,m_3=8kg,u_1=4m/s,u_2=1m/s.$$
Thus, on solving further we get,
$\left(v_2-v_1\right)=4-1=3$
$\left(v_2-v_1\right)=3$ ……………. (1)
Now on considering the law of conservation of linear momentum we get,
Momentum Before collision= momentum after collision
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
Put values from the given data, we get
$2\times 4+4\times 1=2v_1+4v_2$
$v_1+2v_2=6$ ………… (2)
Now From the equation (1) and (2),
We get, $v_1=0m{s}^{-1}\text{ and }{\text{v}}_2=3m{s}^{-1}$
Therefore the correct option is (B).

Additional information:
The strong physical interaction among bodies involving the exchange of momentum in a small interval of time is called a collision. In the collision, particles may or may not come in real physical contact. Based on the conservation of the total kinetic energy of a colliding particle, collisions are classified into two types.
A) Elastic collision
B) Inelastic collision.
The collision, in which both momentum and kinetic energy remain constant or unchanged is known as elastic collisions. In this type of collision, the colliding particle regains their shape and size completely after the collision that is no fraction of mechanical energy is stored as deformation potential energy in the bodies. Examples: the collision between atomic particles, the collision of alpha particles with the nucleus.
The collision, in which kinetic energy is not conserved but the law of conservation of momentum holds good is known as an inelastic collision. However, the total energy is conserved in this type of collision. In an inelastic collision, the colliding particles do not regain their shape and size completely after the collision. Example: the collision between two billiard balls, a collision between two automobiles on a road.
In a collision, the two colliding bodies stick together and move with a common velocity, then the collision is called a perfectly inelastic collision. In a perfectly inelastic collision, linear momentum is conserved, kinetic energy is not conserved. But total energy is conserved. Example: the collision between bullet and block of wood into which it is fired when the bullet remains embedded in the block.

Note:
In this type of question, apply the law of conservation of momentum as well as kinetic energy. And find out final velocities after the collision. Some of the examples of elastic collision are,
*When a billiard ball on a billiard table hits another ball.
*When you hit a ball on the ground, it bounces back to you, you observed there is no net change in the kinetic energy of the ball. It is an elastic collision.