17 Two balls of clay collide, stick together, and move as a single entity. The l

Question

17 Two balls of clay collide, stick together, and move as a single entity. The larger ball of clay has twice the mass of the smaller one, both come into the collision with their velocity vectors making the same angle with the horizontal, and the combined ball of clay continues in the horizontal direction at a speed one-third as great as the incoming speed of the smaller ball of clay. There are no external forces present a. Find the angle 0. b. Find the fraction of the system's initial kinetic energy that is converted into thermal energy as a result of the collision. Assume the combined lump doesn't rotate after the collision.

Explanation / Answer

Initial velocity of mass m1 = v1 ( cos(theta)i - sin(theta)j)
Initial velocity of mass m2 = v2 ( cos(theta)i + sin(theta)j)

initial momentum of the system = m1v1( cos(theta)i _ sin(theta)j) + m2v2 ( cos(theta)i + sin(theta)j)
final momentum of the system = (m1 + m2)vf i

from momentum balance in both directions
m1v1sin(theta) = m2v2sin(theta)
=> m1v1 = m2v2
but m1 = 2m2
so, 2m2*v1 = m2V2
v1 = v2/2

also, vf = v2/3
so, m1v1cos(theta) + m2v2cos(theta) = (M1 + m2)vf
2m2*v2/2 + m2*v2 = (m1 + m2)V2/2cos(theta)
cos(theta) = (3*m2*v2)/2*2m2v2 = 3/4
a. theta = arccos(3/4) = 41.40 deg
b. final energy = 0.5(m1 + m2)vf^2 = 0.5*(3m2)*v2^2/9 = m2*v2^2/6 J
initial energy = m1v1^2 + m2v2^2 /2 = 2m2V2^2/8 + m2v2^2/2 = 3m2v2^2/4 J
energy lost = initial - final = 7m2v2^2/12 J

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