A 0.145 kg baseball is pitched at 42 m/s. The batter hits it horizontally to the pitcher at 58 m/s.?

Question

A 0.145 kg baseball is pitched at 42 m/s. The bater hits it horizontally to the pitcher at 58 m/s.

a. Find the change in momentum of the ball.
b. If the ball and are in contact 4.6*10^-4 seconds, what is the average force during contact?

Please help me with this question.
I'm having trouble understanding it.

Answer 1

A. Change in momentum = initial momentum - final momentum
= (0.145 kg * - 58 m/s ) - (0.42 kg * 42 m/s)
= x kg.m/s

B. Newton's second law.

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Answer 2

Initial momentum = mass x initial velocity = 0.145 x 42 kg-m/s
Final momentum = mass x final velocity = 0.145 x (-58) kg-m/s

Change in momentum
= final momentum - initial momentum
= 0.145 x (-58) kg-m/s - 0.145 x 42 kg-m/s

Average force
= (change in momentum) / time
= [0.145 x (-58) kg-m/s - 0.145 x 42 kg-m/s] / (4.6 x 10^-4) s
=
(kg-m/s^2 is Newton N)

Answer 3

a) Definition of change of momentum = pf - pi = mvf - mvi = m(vf-vi) = 0.145kg ( -58 m/s - 42 m/s) = -100 kg m/s.

b) Newton's second law (F = ma) can be restated as:
Fav * (tf-ti) = pf - pi. To get average force,
Fav = (pf - pi)/(tf - ti) = 217 kN.

Answer 4

a. mass*(v_initial - v_final)

b. Change in velocity divided by time is acceleration.