Question

In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue ball bounces off with a speed of 0.8 m/s at an angle of 20°, as shown in the diagram below. Both balls have a mass of 0.6 kg.

PICTURE

a. What is the momentum of the system before the collision? (Write it in component form.) (0.5 points)
b. What is the momentum of the system after the collision? (0.5 points)
c. Write the velocity of the cue ball after the collision in component form. (1 point)
d. What is the x-component of the velocity of the eight ball after the collision? (1 point)
e. What is the y-component of the velocity of the eight ball after the collision? (1 point)
f. At what angle does the eight ball travel after the collision? (1 point)
g. What is the magnitude of the eight ball's velocity after the collision? (1 point)

Answer 1

a)  p₀ = 1.2 kg m / s

b) p_f = 1.2 kg m / s  

c)    $v_{2f}$ = 1.278 m/s

d)1.2482 =$v_{2f}$ cos θ

e)  0.2736 =$v_{2f}$sin θ

g) θ = 12.36

What is momentum?

Momentum is the product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction.

According to the question,

We define a system formed by the two balls, which are isolated and the forces during the collision are internal.

Hence , the momentum is conserved

a) The momentum of the system before the collision:

$p_o = m v_1_0 + 0 p_o= 0.6 2 p_o = 1.2 kg m / s$

b) The momentum of the system after the collision, as the system is isolated, the momentum is conserved so

$p_f = 1.2 kg m / s$

d. On X axis

$p_ox$ = 1.2 kg m / s

$p_{fx}$= m v1f cos 20 + m v2f cos θ

$p_o = p_f$

1.2 = 0.6 (-0.8) cos 20+ 0.6$v_{2f}$ cos θ

1.2482 =$v_{2f}$ cos θ

e. On Y axis  

$p_{oy} = 0$

$p_{fy}$= m$v_{1f}$sin 20 + m$v_{2f}$cos θ

0 = 0.6 (-0.8) sin 20 + 0.6 $v_{2f}$ sin θ

0.2736 =$v_{2f}$sin θ

Here,

g. we write our system of equations

0.2736 =$v_{2f}$ sin θ

1.2482 = $v_{2f}$ cos θ

0.219 = tan θ (divide)

θ = tan⁻¹ 0.21919

θ = 12.36

c. Now To find the velocity

0.2736 = $v_{2f}$ sin θ

$v_{2f}$= 0.2736 / sin 12.36

$v_{2f}$ = 1.278 m / s

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