When a baseball bat strikes a ball, the impact can be approximately regarded as

Question

When a baseball bat strikes a ball, the impact can be approximately regarded as an elastic collision (the hands of the hitter have little effect on the short time the bat and the ball are in contact). Suppose that a bat of 060 kg moving horizontally at 35 m/s encounters a ball of 015 k9 moving at 35 m/s in the opposite direction. We cannot directly look at this collision as an elastic collision in one dimension, since both particles are in motion before collision (vi = -35 m/s and = 35 m/s). However, we can do so if we use a reference frame that moves at a velocity V0 -35 m/s in the direction of the initial motion of the bat. In this reference frame, the initial velocity of the bat is zero. (a) What are the final velocities of the ball and the bat, just after the collision? Number Bat: (b) What are these final velocities in the reference frame of the ground?

Explanation / Answer

its given that,

mass of bat = m1 = 0.60 kg and v1 = 35m/s

mass of ball =m2 = 0.15 kg and v2 = 35m/s

As per the referance frame, v1 =0

V be the final velocity of the bat,

V = [v1 (m1-m2)+2m2v2]/(m1+m2) = 2m2v2/(m1+m2) = 2 x 0.15 x 35 / (0.60+0.15) = 14m/s

V' be the final velocity of the ball,

V' =[ v2(m2-m1)+2m1v1](m1+m2) = v2(m2-m1) / (m1+m2) = 35 (0.15-0.60)/(0.60+0.15)= 21 m/s

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