UT head baseball coach Dave Serrano throws a 1.9 kg ball at the suspended 26.6 k

Question

UT head baseball coach Dave Serrano throws a 1.9 kg ball at the suspended 26.6 kg block with a velocity of 5.6 m/s. The coefficient of restitution between the ball and the block is e = 0.82. Use right as the positive direction.

What was the percentage of kinetic energy lost during the collision? (%) UT head baseball coach Dave Serrano throws a 1.9 kg ball at the suspended 26.6 kg block with a velocity of 5.6 m/s. The coefficient of restitution between the ball and the block is e = 0.82. Use right as the positive direction. What was the percentage of kinetic energy lost during the collision? (%)

Explanation / Answer

By conservation of momentum,

m1v1i + m2v2i = m1v1f + m2v2f

As v1i = 5.6 m/s, v2i = 0, m1 = 1.9 kg, m2 = 26.6 kg, then

10.64 = 1.9v1f + 26.6v2f

--> 5.6 - 14v2f= v1f       [1]

Using the coefficient of resitution,

0.82(v1i - v2i) = (v2f - v1f)

4.592 = V2f - V1f

Thus,

v1f = v2f - 4.592      [2]

Comparing equations [1] and [2],

v2f - 4.592 = 5.6 - 14v2f

Solving fopr v2f,

v2f = 0.6795 m/s

Using this to solve for v1f in [1].

v1f = -3.913 m/s

Thus, the total KE after the collision is

KE1f + KE2f = 20.69 J

Initially,

KE1i + KE2i = 29.792

Thus, KE lost = 29.792 - 20.69 J = 9.102 J

Thus, as a percentage,

%energy lost = 9.102 J/29.792 J *100% = 30.6%   [ANSWER]

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