Question

The cue ball collides with the 8-ball on a pool table. Before the collision, the 8-ball was stationary, and the cue ball was moving east with a speed of 9 m/s . After the collision, the 8-ball moves at an angle of 45 ∘ south of east with a speed of 4 m/s . Right after the collision, what is the component of the cue ball's velocity that points east? You may assume the two balls have the same mass. Right after the collision, what is the component of the cue ball's velocity that points north?

Answer 1

We know that after the collision, Momentum will remain conserved, So

Using Momentum conservation in x-direction:

Pix = Pfx

m1*u1x + m2*u2x = m1*v1x + m2*v2x

m1 = mass of cue ball = m

m2 = mass of 8-ball = m

u1x = Initial velocity of m1 in x-direction = 9 m/sec

u2x = Initial velocity of m2 in x-direction = 0 m/sec

Suppose after collision m1 is moving with v1 velocity at angle North of east, then

v1x = Final velocity of m1 in x-direction = v1*cos deg

Give after collision m2 is moving with 4 m/sec velocity at angle 45 South of east, then

v2x = Final velocity of m2 in x-direction = 4*cos 45 deg = 2.828 m/sec

So,

m*9 + m*0 = m*v1*cos + m*2.828

9 + 0 = v1*cos + 2.828

v1*cos = component of cue ball's velocity that points towards east = 9 - 2.828 = 6.172 m/sec

Now Using Momentum conservation in y-direction:

Piy = Pfy

m1*u1y + m2*u2y = m1*v1y + m2*v2y

u1y = Initial velocity of m1 in y-direction = 0 m/sec

u2y = Initial velocity of m2 in y-direction = 0 m/sec

Suppose after collision m1 is moving with v1 velocity at angle North of east, then

v1y = Final velocity of m1 in y-direction = v1*sin deg

Give after collision m2 is moving with 4 m/sec velocity at angle 45 South of east, then

v2y = Final velocity of m2 in y-direction = -4*sin 45 deg = -2.828 m/sec

So,

m*0 + m*0 = m*v1*sin + m*(-2.828)

0 + 0 = v1*sin - 2.828

v1*sin = component of cue ball's velocity that points towards north = 2.828 m/sec

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