As per the question the mass of the ball [m]=57 gram
1 gram\(=10^{-3} kg\),hence 57 gram =0.057 kg
The ball was initially moving with velocity of 45 m/s to right.
Hence initial velocity[u] =45 m/s.
After reversing its direction due to the wall,the velocity of the ball was reduced to 33 m/s to the left.
Hence the final velocity of the ball [v] =33 m/s.
The time of contact of the ball with the wall [t] =0.0085 s
we are asked to calculate the force exerted by the wall on the ball.
First we have to calculate the impulse.
The impulse of a force is defined as the change in momentum or the force multiplied by time.
Mathematically impulse=force×time =change in momentum
⇒ F×t =m[v-u]
⇒\(F= m\frac{v-u}{t}\)
=\(0.057*\frac{[-33-45]}{0.0085}\) [V= -33 m/s as it is opposite in direction]
= - 523.0588 N
The answer is very close to -520 N.
Hence option D is right. Here negative sign is due to the fact the force acts to the left i.e opposite to the direction of motion .
