1). A 0.145-kg baseball pitched horizontally at 30.3m/s strikes a bat and is pop

Question

1). A 0.145-kg baseball pitched horizontally at 30.3m/s strikes a bat and is popped straight up to a height of 55.7m.?
A). If the contact time is 2.30m/s , calculate the magnitude of the average force on the ball during the contact.
B).Calculate the direction of the average force on the ball during the contact.

2). Two billiard balls of equal mass undergo a perfectly elastic head-on collision.
If one ball's initial speed was 1.50m/s , and the other's was 3.50m/s in the opposite direction, what will be their speeds after the collision?

Explanation / Answer

1) ball goes vertically 55.7 m high so its veloicty just after strike is v then.
mv^2 / 2 = mgh   ( using work energy theorem)
v = sqrt(2gh) = 33.06 m/s
Impulse = change in momentum
= mv - mu
= 0.145 ( 33.06j - 30.3 i) =   4.79j - 4.39 i   kg.m/s

Impulse = Force x time
4.79j - 4.39 i    = F x 2.30 x 10-3
F = - 1910.22i + 2082.61 j   N

magnitude = sqrt(1910.22^2 + 2082.61^2 ) = 2826 N ......Ans

B ) direction = 180 - tan-1(2082.61 / 1910.22) = 132.53 degrees from initial direction.

2) in perfectly elastic head on collision , if mass of the both obejct are same then
they just exchange their velocities.
one ball's final v = 3.50 m/s in opposite dierction.
second ball;s   = 1.50 m/s

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