A ball moving with speed v hits another identical ball at rest. The two balls stick together after collision. If specific heat of the material of the balls is S, the temperature rise resulting from the collision is

The correct option is v²/4S.

Let us analyze the collision and the subsequent temperature rise step-by-step:

Step 1: Conservation of Linear Momentum
Let the mass of each identical ball be $m$. The first ball moves with a velocity $v$ and hits the second identical ball which is at rest (initial velocity is 0). Since they stick together after the collision, they move with a common final velocity, let's call it $v^{\prime }$.
According to the law of conservation of linear momentum:

$mv+m\left(0\right)=(m+m)v^{\prime }$

$mv=2m{v}^{\prime }$

$v^{\prime }=\frac{v}{2}$

Step 2: Calculate the Loss in Kinetic Energy
The initial kinetic energy of the system ($K_i$) before the collision is:

$K_i=\frac{1}{2}m{v}^2$

The final kinetic energy of the combined system ($K_f$) after the collision is:

$K_f=\frac{1}{2}\left(2m\right){\left(v^{\prime }\right)}^2=m{(\frac{v}{2}}^{})2=\frac{m{v}^2}{4}$

The loss in kinetic energy ($\Delta K$) during the collision is:

$\Delta K=K_i-K_f=\frac{1}{2}m{v}^2-\frac{1}{4}m{v}^2=\frac{1}{4}m{v}^2$

Step 3: Relate the Loss in Kinetic Energy to Temperature Rise
This lost kinetic energy is converted into heat energy ($Q$) which raises the temperature. Considering the heat capacity per unit mass of the colliding ball system, we have:

$Q=mS\Delta T$

Equating the loss in kinetic energy to the heat energy generated:

$\frac{1}{4}m{v}^2=mS\Delta T$

Solving for the rise in temperature ($\Delta T$):

$\Delta T=\frac{v^2}{4S}$