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

Answer:

-2250 N

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