Extension Work on Momentum in Two Dimensions You are nearing the end of a game o
ID: 1783778 • Letter: E
Question
Extension Work on Momentum in Two Dimensions You are nearing the end of a game of pool and only the cue ball and the 8 ball remain. In order not to scratch, you would like to know how the cue ball will deflect if you strike the cue ball such that it moves with a velocity v and it collides with the 8 ball which is initially at rest. Assume that in order to sink the 8 ball for the win you need the 8 ball to emerge from the collision with a velocity u at an angle to the direction the cue ball was initially moving in. Also assume all pool balls have the same mass. View from above: Initially a. With what velocity and at what angle will the cue ball deflect? b. Suppose you strike the cue ball with an initial velocity of magnitude 1.5m/s, and you need the 8 ball to have a final velocity of magnitude 1 m/s at an angle of 309. At what angle will the cue ball deflect?Explanation / Answer
by conservation of momentum
initial horizontal momentum = final horizontal momentum
mv = mu * cos(theta) + mx * cos(alpha)
v = u * cos(theta) + x * cos(alpha)
x * cos(alpha) = v - u * cos(theta) ---------- (1)
initial vertical momentum = final vertical momentum
0 = mu * sin(theta) - mx * sin(alpha)
u * sin(theta) = x * sin(alpha) ------------- (2)
squaring 1 and 2 and adding them we'll get
v^2 + u^2 * cos(theta)^2 - 2uv * cos(theta) + u^2 * sin(theta)^2 = x^2 (sin(alpha)^2 + cos(alpha)^2)
v^2 + u^2 - 2uv * cos(theta) = x^2
x = sqrt(v^2 + u^2 - 2uv * cos(theta))
velocity = sqrt(v^2 + u^2 - 2uv * cos(theta))
angle = sin^-1(u * sin(theta) / x)
angle = sin^-1(u * sin(theta) / sqrt(v^2 + u^2 - 2uv * cos(theta)))
putting values we'll get
velocity = sqrt(1.5^2 + 1^2 - 2 * 1.5 * cos(30))
velocity = 0.807 m/s
angle = sin^-1(1 * sin(30) / 0.807)
angle = 38.29 degree
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