Question

A ball reaches a wall at 60 m/s along +ve X direction, and leaves the wall in the opposite direction with the same speed. Assuming that the mass of the ball as 50gm and the contact time is 0.02 second, the force exerted by the wall on the ball is

• A. 300 N along + X direction
• B. 300 N along - X direction
• C. 3,00,000 N along + X direction
• D. 3,00,000 N along - X direction

Answer 1

The correct option is B 300 N along - X direction

Force exerted by the wall on the ball is the change in momentum of the ball per unit time.
Initial momentum$=mv=50\times {10}^{-3}\times 60=3Ns$
So the momentum just after collision will be $-3Ns$
$\therefore$change is momentum$=3-\left(3\right)=6Ns$

Force$=\frac{6}{0.02}=300N$
The force exerted by the wall on the ball would be opposite to its initial velocity. Thus the direction of force would be negative X direction.