A 0.145 kg baseball pitched at 39.0 m/s is hit on a horizontal line drive straight back toward the pitcher at 52.0 m/s. If the contact time between bat and ball is 1.00\times10^{-3} s,calculate the average force between the bat and ball during contest.

Question

A 0.145 kg baseball pitched at 39.0 m/s is hit on a horizontal line drive straight back toward the pitcher at 52.0 m/s. If the contact time between bat and ball is $1.00 \times 10^{- 3}$ s,calculate the average force between the bat and ball during contest.

Answer 1

This problem is really quite simple. All you need to do is use the

F △ t = m △ v

equation.
First, you are looking for force so rewrite theequation solving for force:

F = \frac{m △ v}{△ t}

Now

△ v

is 39.0 m/s- (-52.0 m/s) = 91 m/s.
52 is negative because the ball ends up traveling in the opposite direction after coming in contact with the ball.
Now just plug in the rest of the information and you get yours answer.

F = \frac{(0.145 k g) (91 \frac{m}{s})}{1 \times 10^{- 3} s} = 13195 N

Answer 2

consider the velocity towards the pitcher as positive

m=mass of the baseball=0.145 kg

v_0 = initial velocity of the baseball = - 39 m/s

v = final velocity of the baseball = 52 m/s

t = time of contact = $1.00 \times 10^{- 3}$ sec

F = average force between bat and ball

Using impulse-change in momentum equation

$F t = m (v - v_{0})$

$F (1 \times 10^{- 3}) = (0.145) (52 - (- 39))$

F=13195N

A 0.145 kg baseball pitched at 39.0 m/s is hit on a horizontal line drive straight back toward the pitcher at 52.0 m/s. If the contact time between bat and ball is 1.00\times10^{-3} s,calculate the average force between the bat and ball during contest.

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