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Question

A 0.150 kg baseball pitched at 40.0 m/s is hit on a horizontal line drive straight back at the pitcher at 50.0 m/s.

a) If the contact time between bat and ball is 3.90 ×10⁻³ s, calculate the magnitude of the force (assumed to be constant) between the ball and bat.  

b)  Determine the direction of the force exerted on a ball.

Answer 1

a) If the contact time between bat and ball is 3.90 ×10^-3 s, calculate the magnitude of the force (assumed to be constant) between the ball and bat.

The pitched ball has an initial momentum p_i = m × v_i, where m is the mass of the ball, and v_i is the velocity of the pitched ball. After the ball is hit, its final velocity is v_f, so its final momentum p_f = m x v_f.

The magnitude of the force acting on the ball is the change in momentum divided by the duration of contact Δt:

F = (p_f-p_i)/Δt

Plugging in the known values:

m = 0.15 kg

v_i = 40 m/s

v_f = -50. m/s

F = 0.15 × (-50.-40.)/3.90 ×10^-3 kgm/s² = - 3462. kgm/s² = - 3462. N

b)  Determine the direction of the force exerted on a ball.

The sign of the Force is negative, meaning it is in the direction opposite to the velocity of the ball after it was pitched.