A super-ball is to bounce elastically back and forth between two rigid walls at a distance from each other. Neglecting gravity and assuming the velocity of super-ball to be \[{v_0}\] horizontally, the average force (in large time interval) being exerted by the super ball on one wall is:
A.\[\dfrac{1}{2}\dfrac{{mv_0^2}}{d}\]
B.\[\dfrac{{mv_0^2}}{d}\]
C.\[\dfrac{{2mv_0^2}}{d}\]
D.\[\dfrac{{4mv_0^2}}{d}\]

Hint: Use the formula for the momentum of an object and determine the change in momentum of the super-ball. Then determine the time required for the super-ball to bounce between two walls once. Use the formula for force in terms of change in momentum to determine the average force exerted by the super-ball on the wall.

Formula used:
The momentum \[P\] of an object is
\[P = mv\] …… (1)
Here, \[m\] is the mass of the object and \[v\] is the velocity of the object.
The speed \[v\] of an object is
\[v = \dfrac{x}{t}\] …… (2)
Here, \[x\] is the distance travelled by the object and \[t\] is the time.
The change in momentum \[\Delta P\] of an object is
\[F = \dfrac{{\Delta P}}{t}\] …… (3)
Here, \[F\] is the force exerted by the object and \[t\] is time.

Complete step by step answer:
We have given that a super-ball is bouncing between two rigid walls with a horizontal velocity \[{v_0}\].
Let us determine the change in momentum of the ball when it bounces once between the rigid walls.
Let the super-ball strike the wall in the left hand side with a velocity \[ - {v_0}\] as the velocity is in the negative X-direction.
According to equation (1), the initial momentum of the ball is
\[{P_i} = - m{v_0}\]
We can say that now the velocity of the ball while striking the opposite wall will be \[{v_0}\].
According to equation (1), the final momentum of the ball is
\[{P_f} = m{v_0}\]
The change in momentum of the super-ball is given by
\[\Delta P = {P_f} - {P_i}\]
Substitute \[m{v_0}\] for \[{P_f}\] and \[ - m{v_0}\] for \[{P_i}\] in the above equation.
\[\Delta P = \left( {m{v_0}} \right) - \left( { - m{v_0}} \right)\]
\[ \Rightarrow \Delta P = 2m{v_0}\]
Let us now determine the average time required for the ball to come back and hit the right side wall again.
Rearrange equation (2) for the time \[t\].
\[t = \dfrac{x}{v}\]
The distance travelled by the ball for the second hit is \[2d\].
Substitute \[2d\] for \[x\] and \[{v_0}\] for \[v\] in the above equation.
\[ \Rightarrow t = \dfrac{{2d}}{{{v_0}}}\]
We can now determine the force exerted by the super-ball on the wall.
Substitute \[2m{v_0}\] for \[\Delta P\] and \[\dfrac{{2d}}{{{v_0}}}\] for \[t\] in equation (3).
\[ \Rightarrow F = \dfrac{{2m{v_0}}}{{\dfrac{{2d}}{{{v_0}}}}}\]
\[ \Rightarrow F = \dfrac{{mv_0^2}}{d}\]
Therefore, the average force exerted by the super-ball on the wall is \[\dfrac{{mv_0^2}}{d}\].

So, the correct answer is “Option B”.

Note:
 One can consider any direction of the motion of the ball between the two walls as positive and negative. If we consider the directions of X-axis opposite to that used in the above solution then the momentum of the ball will be negative in sign. Also the time required for one bounce between two walls comes out with a negative sign. Hence, the final answer will be the same.