Question

"A 0.15 kg ball moving at 40 m/s is struck by a bat. The bat reverses the ball's direction and gives it a speed of 50 m/s. What average force does the bat apply to the ball if they are in contact for 6.0 ×10 -3 s?"

Answer 1

The average force exerted by the bat on the ball is 2250 N.

Step-by-step explanation:

To calculate the average force exerted by the bat on the ball, we can use the impulse-momentum principle. The impulse experienced by the ball is equal to the change in its momentum. We can calculate the initial momentum by multiplying the mass of the ball by its initial velocity, and similarly, we can calculate the final momentum using the mass and final velocity. By subtracting the initial momentum from the final momentum, we get the change in momentum. Finally, dividing the change in momentum by the time of contact gives us the average force.

Using the given values, the initial momentum of the ball is (0.15 kg) × (-40 m/s) = -6 kg·m/s, and the final momentum is (0.15 kg) × (50 m/s) = 7.5 kg·m/s. The change in momentum is 7.5 kg·m/s - (-6 kg·m/s) = 13.5 kg·m/s. Dividing this by the time of contact, 6.0 × 10^-3 s, gives us an average force of 2250 N.

Answer 2

-2250 N

Step-by-step explanation:

From the question,

Using.

F = m(v-u)/t .................. Equation 1

Where F = Average force the bat apply to the ball, m = mass of the ball, v = final velocity of the ball, u = initial velocity of the ball, t = time.

Given: m = 0.15 kg, v = -50 m/s(reversed direction), u = 40 m/s, t = 6.0×10⁻³ s = 0.006 s.

Substitute into equation 1

F = 0.15(-50-40)/0.006

F = 0.15(-90)/0.006

F = -2250 N.

The force is negative because it opposes the initial motion of the ball.

Hence the bat average force = -2250 N