A pool player hits the cue ball giving it a speed of 5.57 m/s and directs its ce
ID: 3163382 • Letter: A
Question
A pool player hits the cue ball giving it a speed of 5.57 m/s and directs its center on a path tangent to the surface of the target ball having the same mass as the cue ball. After the collision the speed of the initially-stationary ball moves with a speed of 4.82 m/s. After the collision, find the new speed and direction of the cue ball. A block of mass m = 4.8 kg, moving to the right on a horizontal frictionless surface with a speed of 2.4 m/s, makes a perfectly elastic collision with a black of mass M at rest. After the collision, the 4.8 kg block recoils with a speed of 0.200 m/s. If the blocks are in contact for 0.17 seconds, determine the magnitude of the average force on the 4.8 kg block while the two blocks are in contact. The angular acceleration of a wheel is given in rad/s^2 by 54t^3 - 8t^4, where t is in seconds. If the wheel starts from rest at t = 0.00 seconds, find the next time the wheel is at rest.Explanation / Answer
A pool player is attempting a fancy shot. He hits the cue ball giving it a speed of 5.57 m/s and directs its center on a path tangent to the surface of the target ball having the same mass as the cue ball. After the collision (on a frictionless tables the initially-stationary ball moves with a speed of 4.82 m/s. After the collision, the new speed of the cue ball.
The direction is the first thing.If the surface of the ball is hard, shiny and slippery then there can be no forces tangential to the surface of the balls at the moment that they hit.
So any change of momentum MUST lie along the line connecting the two centres at this moment.
In this problem there is one other rule that has been used.
Energy is conserved. This is not true in a real game but as they have not given a direction then we have to assume something to solve the problem.
hence 1/2 * m * 5.57^2 = 1/2 * m * 4.82^2 + 1/2 * m * v^2
divide both sides by 1/2 m and solve for v
A block of mass m = 4.8 kg, moving on a horizontal frictionless surface with a speed 2.4 m/s, makes a perfectly elastic collision with a block of mass M at rest. After the collision, the 4.8 kg block recoils with a speed of 0.200 m/s. In the figure, the blocks are in contact for 0.17 s. The magnitude of the average force on the 4.8-kg block, while the two blocks are in contact.
F = mv / t
F = 4.8(0.200 - (-2.4)) / (0.17 - 0.000)
F = -73.41
F = 73.41 N
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