As per the question the mass of the ball [m]=57 gram
1 gram[tex]=10^{-3} kg[/tex],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]
⇒[tex]F= m\frac{v-u}{t}[/tex]
=[tex]0.057*\frac{[-33-45]}{0.0085}[/tex] [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 .
Answer:
Force, F = 520 N
Explanation:
It is given that,
Mass of the tennis ball, m = 57 g = 0.057 kg
Initial speed of the ball, u = 45 m/s
Final speed of the ball, v = -33 m/s
Time of contact, t = 0.0085 s
Let F is the magnitude of the force that was exerted on the ball. It is equal to :
[tex]F=m\times \dfrac{v-u}{t}[/tex]
[tex]F=0.057\ kg\times \dfrac{-33\ m/s-45\ m/s}{0.0085\ s}[/tex]
F = −523.05 N
Out of given options, the magnitude of force exerted on the ball is 520 N and it is acting in opposite direction.
