A ball with a mass of 0.625 kg is initially at rest. It is struck by a second ba

Question

A ball with a mass of 0.625 kg is initially at rest. It is struck by a second ball having a mass of 0.425 kg , initially moving with a velocity of 0.240 m/s toward the right along the axis. After the collision, the 0.425 kg ball has a velocity of 0.170 m/s at an angle of 36.6 degrees above the axis in the first quadrant. Both balls move on a frictionless, horizontal surface.
Part A: What is the magnitude of the velocity of the 0.625 kg ball after the collision?
Part B: What is the direction of the velocity of the 0.625 kg ball after the collision?
Part C: What is the change in the total kinetic energy of the two balls as a result of the collision?

Explanation / Answer

1. 0.625 2. -10.9° with original direction 3. 0.117 Let v = velocity of ball 0.600 after collision at an angle ?. Balancing momentum along horizontal direction (0.400) * (0.25) = (0.20) cos (36.9°) + v cos ? => v cos ? = 0.625 Balancing momentum along vertical direction 0 = (0.20) sin (36.9°) - v sin ? => v sin ? = 0.120 => tan ? = 0.120 / 0.625 => ? = 10.9° cos 10.9° = 0.982 => v = 0.625/0.982 = 0.636 Change in K.E. = Final K.E. - initial K.E. = (1/2) [ (0.400) * (0.2)^2 + (0.600) * (.636)^2 - (0.400) * (0.25)^2 ] = (0.5) [ 0.016 + 0.243 - 0.025 ] = 0.117

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