Question

A baseball player swings his 1kg bat at 10m/s as he hits a 60g baseball that is coming at him at 20m/s. The ball bounces of his racket at 38m/s.

(a) Draw a before/after picture showing velocities/directions.

(b) How fast is his bat moving after the collision? In what direction? If the bat and ball are in contact for only 10ms, what is the average force on the ball during the collision?

Answer #1

a)

b)

m_{B} = mass of bat = 1 kg

m_{b} = mass of ball = 60 g = 0.060 kg

using conservation of momentum

m_{B} v_{Bi} + m_{b} v_{bi} =
m_{B} v_{Bf} + m_{b} v_{bf}

(1) (10) + (0.060) (- 20) = (1) v_{Bf} + (0.060)
(38)

v_{Bf} = 6.5 m/s

in the same direction as the bat was moving before collision .

F_{avg} = average force

using impulse-change in momentum equation for bat

F_{avg} t = m_{B} (v_{Bf} -
v_{Bi})

F_{avg} (10 x 10^{-3}) = (1) (6.5 - 10)

F_{avg} = - 350 N

negative sign indicates left direction

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