A ball with a mass of 0.620 kg is initially at rest. It is struck by a second ba

Question

A ball with a mass of 0.620 kg is initially at rest. It is struck by a second ball having a mass of 0.370 kg , initially moving with a velocity of 0.245 m/s toward the right along the x axis. After the collision, the 0.370 kg ball has a velocity of 0.195 m/s at an angle of 36.3 above the x axis in the first quadrant. Both balls move on a frictionless, horizontal surface. 1) What is the magnitude of the velocity of the 0.620 kg ball after the collision? 2)What is the direction of the velocity of the 0.620 kg ball after the collision? 3)What is the change in the total kinetic energy of the two balls as a result of the collision? A ball with a mass of 0.620 kg is initially at rest. It is struck by a second ball having a mass of 0.370 kg , initially moving with a velocity of 0.245 m/s toward the right along the x axis. After the collision, the 0.370 kg ball has a velocity of 0.195 m/s at an angle of 36.3 above the x axis in the first quadrant. Both balls move on a frictionless, horizontal surface. 1) What is the magnitude of the velocity of the 0.620 kg ball after the collision? 2)What is the direction of the velocity of the 0.620 kg ball after the collision? 3)What is the change in the total kinetic energy of the two balls as a result of the collision?

Explanation / Answer

1) .37*.245 = .620*vcos(theta) + .37*.195*cos36.3

also:

0 = .62*vsin(theta) + .37*.195*sin(36.3)

from two equation get v and theta,

for final part:

change in KE

change = (.5*.37*2.45^2) - ( .5*.37*.195^2 + .5*.62*v^2 )

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