Question

A ball (0.280 kg) approaches a player horizontally at v = 15 m/s. The player takes the ball and causes it to move in the opposite direction with v = 22 m/s. Calculate:

a) Impulse delivered by the player to the ball
b) If the player's fist is in contact with the ball for 0.06 s, calculate the average force exerted on the player's fist.
a) 1.12 N∙s, 18.67 N
b) 1.68 N∙s, 28 N
c) 2.04 N∙s, 34 N
d) 2.40 N∙s, 40 N

Answer 1

The impulse delivered by the player to the ball is 10.36 N•s, and the average force exerted on the player's fist is 172.67 N. The initial and final velocities were accounted for with their direction, resulting in the impulse and force calculated.

Step-by-step explanation:

To calculate the impulse delivered by the player to the ball, you need to use the formula for impulse: Impulse = Δp = mvf - mvi, where m is the mass of the ball, vf is the final velocity, and vi is the initial velocity. The direction of the impulse is important because the final and initial velocities are in opposite directions, so they should be treated algebraically with their signs.

For the case given:

• Mass (m) = 0.280 kg
• Initial velocity (vi) = -15 m/s (negative because the final velocity will be in the opposite direction)
• Final velocity (vf) = +22 m/s

Impulse = (0.280 kg)(22 m/s) - (0.280 kg)(-15 m/s) = 10.36 kg•m/s

The average force exerted on the player's fist can be calculated using the formula for force: F = Impulse/Δt, where Δt is the time interval.

Given:

• Impulse = 10.36 kg•m/s
• Δt = 0.06 s

Average Force = 10.36 kg•m/s / 0.06 s = 172.67 N

Therefore, the correct option is (d) 10.36 N•s, 172.67 N.