A 57 g tennis ball is traveling at 45 m/s to the right, when it hits a racket. The ball reverses direction and travels at 33 m/s. The ball is in contact with the racket for 0.0085 s.
Answers
Answer:
As per the question the mass of the ball [m]=57 gram
1 gram=10^{-3} kg=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}F=m
t
v−u
=0.057*\frac{[-33-45]}{0.0085}0.057∗
0.0085
[−33−45]
[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 .
Explanation:
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