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A 0.130kg baseball pitched at 35.5 m/s is hit on a horizontal line drive straight back at the pitcher at 50.5 m/s .
Part a: If the contact time between bat and ball is 4.25×10^−3 , calculate the magnitude of the force (assumed to be constant) between the ball and bat.

part b)Determine the direction of the force exerted on a ball.
a)the force is directed towards the pitcher
b) the force is directed away from the pitcher.

this exercise is conservation of linear momentum.
initial momentum: (0.130kg)(35.5)(i)
"remember linear momentum is vectorial.
i is vector direction and only the movement is in X"
= 4.615(i)

Final momentum is :
= (0.130)(50.5)(-i)
=6.565(-i)

then I=impulse

I= 6.565(-i) - 4.615(i)
I= -6.565(i) - 4.615(i)
I= -11.18(i)
I= 11.18(-i) Kgm/s

I=F.t
F=I/t= 11.18(-i)Kgm/s/4.25×10^−3 s
F=2630.6 (Kgm/s² o N)

good luck

the magnitude of the force is 2630.6 N

thank you !

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