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. what is the magnitude of the force that was exerted on the ball? a-1.8 n b-12 n c-81 n d-520 n

As per the question the mass of the ball [m]=57 gram

1 gram,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]

                                              ⇒

                                                           = [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 .

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 :

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.

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