A ball with a mass of 0.580 kg is initially at rest. It is struck by a second ba

Question

A ball with a mass of 0.580 kg is initially at rest. It is struck by a second ball having a mass of 0.400 kg , initially moving with a velocity of 0.235 m/s toward the right along the x axis. After the collision, the 0.400 kg ball has a velocity of 0.185 m/s at an angle of 36.7 above the x axis in the first quadrant. Both balls move on a frictionless, horizontal surface.

What is the magnitude of the velocity of the 0.580 kg ball after the collision?

What is the direction of the velocity of the 0.580 kg ball after the collision?

What is the change in the total kinetic energy of the two balls as a result of the collision?

Explanation / Answer

velocity of 0.580 kg ball = 0

initial velocity of 0.400 ball = 0.235 i m/s

final velocity of 0.400 kg = 0.185 ( cos36.7i + sin36.7 j) = 0.148 i + 0.110 j m/s

using momentum conservation,

0.580 x 0 + 0.400 x 0.235i = 0.400( 0.148i + 0.110j) + 0.580v

v = 0.0348 i - 0.044j / 0.580

v =0.06i - 0.0759j

magnitude of velocity = sqrt(0.06^2 + 0.0759^2) = 0.0968 m/s

direction = - tan^-1( 0.0759/0.06) = - 51.67 deg

change in KE = 0.580 x 0.0968^2 /2 + 0.400 x 0.185^2 /2 - 0.400 x 0.235^2 /2 = - 1.48 x 10^-3 J

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