Q. 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
A) $\frac{v^{2}}{8S}$
B) $\frac{v^{2}}{4S}$
C) $\frac{v^{2}}{2S}$
D) $\frac{v^{2}}{S}$
Explanation
(a) Initial momentum $= mv$
Final momentum $= 2mV$
By the conservation of momentum, $mv = 2mV$
$\Rightarrow V = \frac{v}{2}$
K.E. of the system after the collision $= \frac{1}{2}(2m)\left( \frac{v}{2} \right)^{2}$
$\therefore$ loss in K.E.$\ = \frac{1}{2}\ mv^{2} - \frac{1}{4}mv^{2} = \frac{1}{4}mv^{2}$
This loss in K.E. will increase the temperature
$\therefore 2m \times s \times \mathrm{\Delta}t = \frac{1}{4}mv^{2} \Rightarrow \mathrm{\Delta}t = \frac{v^{2}}{8s}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.