A ball with a mass of 0.630 kg is initially at rest. It is struck by a second ba

Question

A ball with a mass of 0.630 kg is initially at rest. It is struck by a second ball having a mass of 0.420 kg , initially moving with a velocity of 0.255 m/s toward the right along the x axis. After the collision, the 0.420 kg ball has a velocity of 0.205 m/s at an angle of 35.9 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 0.630 kg ball after the collision?

What is the direction of the velocity of the 0.630 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

Let v = velocity of ball 0.630 after collision at an angle .
Balancing momentum along horizontal direction
(0.420) * (0.255) = (0.205) cos (35.9°) + v cos
=> v cos = -0.05896

Balancing momentum along vertical direction
0 = (0.205) sin (35.9°) - v sin ======> v sin = 0.120

=> tan = 0.120 / -0.05896 => = -63.87°
cos -63.87° = 0.44
=> v = -0.05896/0.44 = -0.134 m/s
Change in K.E. = Final K.E. - initial K.E.
= (1/2) [ (0.420) * (0.205)^2 + (0.630) * (0.134)^2 - (0.420) * (0.255)^2 ]
= (0.5) [ 0.0177 + 0.0113 - 0.0273 ]
= 0.00085

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