Question

When a 0.058 kg tennis ball is served, it accelerates from rest to a speed of 45.0 m/s. The impact with the racket gives the ball a constant acceleration over a distance of 0.440 m. What is the magnitude of the net force acting on the ball?

a.What is the average velocity of the ball?

b.Over what time period was the ball struck?

c. What is the acceleration of the ball?

d.What is the net force on the ball?

e.If the ball had been heavier, but experienced the same change in velocity, would the applied force have to be greater or lesser than before?

Answer 1

Step-by-step explanation:

Note : the questions do not come in the right order.

Data:

Mass = 0.0518kg

Velocity = 45 m / s

Distance (s) = 0.44m

C) what's the acceleration on the ball?

Using equation of motion,

V² = u² + 2as

V² - u² = 2as

a = (v² - u²) / 2s

a = (45² - 0) / 2 * 0.44 [the ball was at rest]

a = 2025 / 0.88

a = 2301.136m/s²

D) The net force on the ball?

Force = mass * acceleration

F = m*a

F = 0.0518 * 2301.136

F = 119.199N

The force acting on the ball was 133.465N

F = 133.47N

b) time period the ball was struck.

From the relationship between impulse and momentum,

Ft = m * v

133.47 * t = 0.058 * 45

t = 2.61 / 133.47

t = 0.01955s

a) average velocity (V) = total distance covered / total time taken

V = s / t

V = 0.44 / 0.01955

V = 22.50m/s.

e) if the ball was heavier and still experienced the same velocity, the applied would've been lesser than before.