Question

If $n$ balls hit elastically and normally on a surface per unit time and all the balls have mass $m$ moving with the same velocity $u$ then force on the surface is:
A. $mun$
B. $2mun$
C. $\dfrac{1}{2}m{u^2}n$
D. $m{u^2}n$

Answer 1

Hint:An elastic collision is one in which there is no net loss in kinetic energy and momentum remains conserved. Force on an object due to its momentum is defined as $\text{Force} = \dfrac{\text{momentum}}{\text{time}}$ and its SI unit is $Kgm{s^{ - 2}}$ .

Complete step by step answer:
Let us consider these $n$ balls moving with a velocity of $u$ and each having mass of $m$.Momentum of one ball while hitting on the surface is given by $p = mu$. Momentum of one ball while rebounding the surface with same velocity due to elastic collision is given by $p' = - mu$ in the opposite direction. Net change in momentum is calculated as $p - p' = mu + mu$
$p - p' = 2mu$
Since, the total number of balls hitting and rebounding the surface is given as $n$.
So, total change in momentum is $ = 2mun$.
Now, we need to calculate total change in momentum in one second
Total change in momentum in one second is $ = 2mun$ $kgm{s^{ - 2}}$
We know, force is equal to rate of change of momentum with respect to time, so Force is given by $Force = 2mun$ $kgm{s^{ - 2}}$.

Hence, the correct option is B.

Note: In elastic collision initial momentum of a system and final momentum of system remains conserved, and it’s known as principle of conservation of linear momentum. Net Kinetic energy also remains constant. Force on an object is equal to rate of change of momentum which is known as Newton’s second law of motion.