Question

A 0.065-kg tennis ball moving to the right with a speed of 15 m/s is struck by a tennis racket, causing it to move to the left with a speed of 15 m/s. if the ball remains in contact with the racquet for 0.020 s, what is the magnitude of the average force exerted on the ball?

Answer 1

Using the formula for impulse, we can solve this problem.
In terms of change in momentum, impulse is:
I = m (v2 - v1)

Or in terms of force and contact time:
I = F t

Equating the two equations:
m (v2 - v1) = F t

Setting the right direction as positive and the left as negative, we can then substitute values with their respective signs:
0.065 (-15 -15) = F (0.020)
F = -97.5 N

The negative sign indicates that the force was directed to the left.

Answer 2

Force is defined as the push or pull applied to the body. The magnitude of the average force exerted on the ball will be -97.5 N. Where -ve shows that the direction is opposite.

What is force?

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

The given data in the problem is given by

m is the mass of tennis ball=  0.065-kg

u is the speed of ball=15 m/sec

v is the rebound velocity= - 15 m/sec

T is the time of contact=0.020sec

F is the average force=?

From the definition of impulse we know

impulse = force × time

I=Ft

From the definition of impulsive force

$\rm I=\frac{m(v-u)}{} \\\\ \frac{m(v-u)}{}= \rm Ft \rm \\\\ \rm F= \frac{m(v-u)}{t} \\\\ \rm F= \frac{0.065(-15-15)}{0.020} \\\\ \RM F= -97.5 N$

Hence the average force exerted on the ball will be -97.5 N. Where -ve shows that the direction is opposite.

To learn more about the force refer to the link;

https://brainly.com/question/26115859