Question

A rigid ball of mass m strikes a rigid wall at ${60^ \circ }$ and gets reflected without loss of speed as shown in the figure. The value of impulse imparted by the wall on the ball will be:
(A) $mV$
(B) $2mV$
(C) $\dfrac{{mV}}{2}$
(D) $\dfrac{{mV}}{3}$

Answer 1

Hint: This question is related to the topics of work, energy, power and Impulse. Impulse is the integral of a force, F, over the time interval, t, for which it acts. Since force is a vector quantity, impulse is also a vector quantity. Impulse applied to an object produces an equivalent vector change in its linear momentum, also in the resultant direction.

Complete step by step answer:
Here, we are given that a rigid ball of mass m strikes a rigid wall at ${60^ \circ }$ and gets reflected without loss of speed. So, we need to calculate the impulse imparted by the wall on the ball.
Now, we know that impulse can also be defined as the change in momentum.
So, impulse = final momentum – initial momentum
$I = {P_f} - {P_i}$
Initial momentum of the ball perpendicular to the wall,
${P_i} = m( - v\cos {60^ \circ })$
${P_i} = \dfrac{{ - mv}}{2}$
Similarly, Final momentum of the ball perpendicular to the wall,
${P_f} = m(v\cos {60^ \circ })$
${P_f} = \dfrac{{mv}}{2}$
Also, impulse = final momentum – initial momentum
$I = {P_f} - {P_i}$
$I = \dfrac{{mv}}{2} - \left( {\dfrac{{ - mv}}{2}} \right)$
$I = \dfrac{{mv}}{2} + \dfrac{{mv}}{2}$
$I = mv$
$I = mv$
Therefore, The value of impulse imparted by the wall on the ball will be $mV$. So, option (A) is correct.

Note:
The impulse-momentum theorem states that the impulse applied to an object will be equal to the change in its momentum. In this theorem, we see that how a small force applied over a long period of time can be used to produce the small velocity change as a large force applied over a short period of time.