Question

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

Answer 1

In order to answer this question, after observing the question, we will conclude that in this question, Newton's second law of motion is applying, so we will apply the formula of Net force in terms of the rate of change in momentum. As the ball is falling towards the surface, the initial momentum is in minus sign.

Complete step by step answer:
Since the ball bounces back at the same velocity but in the opposite direction,
So, here Newton's second law of motion is applied, i.e..
$\therefore {F_{net}} = n.\dfrac{{dp}}{{dt}}$ …….(i)
where, ${F_{net}}$ is the net force and
$\dfrac{{dp}}{{dt}}$ is the change in momentum, $\Delta p$
So, change in linear momentum of one ball:
$\therefore \Delta p = {p_f} - {p_i}$ …..(ii)
where, ${p_i}$ is the initial momentum and,
${p_f}$ is the final momentum.
As the ball is moving downward, so the initial momentum will be:-
${p_i} = - mu$
where, $m$ is the mass of the ball and $u$ is the velocity of the ball.
and, minus sign indicates that the ball is moving downwards.
And the final momentum will be:
${p_f} = mu$
Now, we will put the value of initial and final momentum in equation(ii):-
$\Rightarrow \Delta p = mu - ( - mu) = 2mu$
Now, we will substitute the value of rate of change in momentum in equation(i), we get:-
$\Rightarrow {F_{net}} = n.2mu = 2m \times u \times n$
Therefore, the force on the surface is $2 \times m \times u \times n$ .
Hence, the correct option is (B) $2 \times m \times u \times n$ .

Note:
When an object moves at twice its normal speed, its momentum doubles. Force causes the object's momentum to change when its speed is changed. As a result, momentum is the product of force and time. However, if you keep exerting force for a certain period of time, the momentum will alter. It is based on Newton's third law of motion.