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 0.245 m/s toward the right along the x axis. After the collision, the 0.400 kg ball has a velocity of 0.210 m/s at an angle of 35.9 above the x axis in the first quadrant. Both balls move on a frictionless, horizontal surface. aWhat is the magnitude of the velocity of the 0.580 kg ball after the collision? b) What is the direction of the velocity of the 0.580 kg ball after the collision? c) 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 0.580 after collision at an angle .

Balancing momentum along horizontal direction
(0.400) * ( 0.245) = (0.21) cos (35.9°) + v cos
=> v cos = 0.576

Balancing momentum along vertical direction
0 = (0.21) sin (35.9°) - v sin
=> v sin = 0.123

=> tan = 0.123 / 0.576 => =12.05°
cos 12.05° = 1.02
=> v = 0.576/ 1.02 = 0.564
Change in K.E. = Final K.E. - initial K.E.
= (1/2) [ (0.400) * (0.21)^2 + (0.580) * (0.564)^2 - (0.400) * (0.245)^2 ]
= 0.169

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