Question

A tennis player practices against a backboard, hitting the ball to the board with a speed of 22 m/s. The ball bounces straight back from the board with a speed of 19 m/s. The mass of the ball is 0.055 kg. What is the average force the wall exerts on the ball if the ball is in contact with the board for 0.0125 s?

Answer 1

The average force the wall exerts on the ball is 180.4 N

Data obtained from the question

From the question given above, the following data were obtained:

• Initial velocity (u) = 22 m/s
• Final velocity (v) = 19 m/s
• Mass (m) = 0.055 kg
• Time (t) = 0.0125 s

• Force (F) =?

Determination of the impulse

We'll begin by calculating the impulse. This can be obtained as follow:

• Initial velocity (u) = 22 m/s
• Final velocity (v) = 19 m/s
• Mass (m) = 0.055 kg
• Impulse (I) =?

I = m(v + u)

I = 0.055 (19 + 22)

I = 0.055 × 41

I = 2.255 Ns

How to determine the force

• Time (t) = 0.0125 s
• Impulse (I) = 2.255 Ns
• Force (F) =?

I = Ft

2.255 = F × 0.0125

Divide both side by 0.0125

F = 2.255 / 0.0125

F = 180.4 N

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