A billiard player sends the cue ball toward a group of three balls that are init

Question

A billiard player sends the cue ball toward a group of three balls that are initially at rest and in contact with one another. After the cue ball strikes the group, the four balls scatter, each traveling in a different direction with different speeds as shown in the figure below. If each ball has the same mass, 0.16 kg,determine the total momentum of the system consisting of the four balls immediately after the collision. (Assume v1 = 0.33 m/s, 1 = 70°, v2 = 0.55 m/s,2 = 30°, v3 = 0.23 m/s, v4 = 0.48 m/s.)

Explanation / Answer

Mass of ball = 0.16 kg
v1 = 0.33 m/s,
1 = 70°,
v2 = 0.55 m/s,
2 = 30°,
v3 = 0.23 m/s,
v4 = 0.48 m/s

X component of Final Momentum = m*v2*cos(30) + m*v1cos(70) - m*v4
X component of Final Momentum = 0.16*0.55*cos(30) + 0.16 * 0.33 *cos(70) - 0.16*0.48 Kg m/s
X component of Final Momentum = 0.01746 Kg m/s

Y component of Final Momentum = m*v3 + m*v2*sin(30) - m*v1*sin(70)
Y component of Final Momentum = 0.16*0.23 + 0.16*0.55*sin(30) - 0.16*0.33*sin(70) Kg m/s
Y component of Final Momentum = 0.0312 Kg m/s

Magnitude of Momentum, P = sqrt(0.01746^2 + 0.0312^2 ) Kg m/s
P =  0.0357 Kg m/s

Direction = 360o - tan^-1(0.0312/0.01746)
Direction = 360o - 60.77o
Direction = 299.23o counterclockwise from the +x axis

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