A cue ball traveling at 4.0 m/s makes a glancing, elastic collision with a target ball of equal mass that is initially at rest. The cue ball is deflected so that it makes an angle of 30° with its original direction of travel.
(a) Find the angle between the velocity vectors of the two balls after the collision.
(b) Find the speed of each ball after the collision.
cue ball m/s
target ball m/s
(a) The angle between the velocity vectors of the two balls after the collision is 120°.
(b) The speed of the cue ball after the collision is 2.0 m/s. The speed of the target ball after the collision is also 2.0 m/s.
To solve this problem, we can apply the law of conservation of momentum and the law of conservation of kinetic energy. Let's break down the steps to find the solutions for each part:
(a) Find the angle between the velocity vectors of the two balls after the collision:
1. Determine the initial and final momenta of each ball.
- Initial momentum of the cue ball (p1_initial) = mass_cue_ball * velocity_cue_ball
- Initial momentum of the target ball (p2_initial) = mass_target_ball * 0 (since it's initially at rest)
- Final momentum of the cue ball (p1_final) = mass_cue_ball * velocity_cue_ball_after_collision (To be determined)
- Final momentum of the target ball (p2_final) = mass_target_ball * velocity_target_ball_after_collision (To be determined)
2. Apply the law of conservation of momentum.
- According to this law, the total momentum before the collision is equal to the total momentum after the collision.
- Therefore, p1_initial + p2_initial = p1_final + p2_final
3. Use the information given to calculate p1_final and p2_final.
- p1_final = mass_cue_ball * velocity_cue_ball_after_collision
- p2_final = mass_target_ball * velocity_target_ball_after_collision
4. Determine the angle between the velocity vectors of the two balls after the collision.
- Use the trigonometric relationship between angles and vectors to find the angle between the velocity vectors of the two balls.
(b) Find the speed of each ball after the collision:
1. Apply the law of conservation of kinetic energy.
- According to this law, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
- Therefore, (1/2) * mass_cue_ball * velocity_cue_ball^2 = (1/2) * mass_cue_ball * velocity_cue_ball_after_collision^2 + (1/2) * mass_target_ball * velocity_target_ball_after_collision^2
2. Use the information given and the result from part (a) to find the velocities of each ball after the collision.
To find the angle between the velocity vectors of the two balls after the collision, we can use the law of conservation of momentum and the law of conservation of kinetic energy.
(a) Conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision:
m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f
Where:
m1 and m2 are the masses of the cue ball and the target ball (equal in this case)
v1i and v2i are the initial velocities of the cue ball and the target ball
v1f and v2f are the final velocities of the cue ball and the target ball
Given:
m1 = m2 (equal masses)
v1i = 4.0 m/s (initial velocity of the cue ball)
v2i = 0 m/s (initial velocity of the target ball)
Let's assume that the final velocities are represented as v1f and v2f, and the angle between them is θ.
Since the target ball is initially at rest, its final velocity is v2f = 0 m/s.
The equation for momentum conservation becomes:
m1 * v1i = m1 * v1f
Simplifying this equation, we get:
v1i = v1f
Therefore, the angle between the velocity vectors of the two balls after the collision is 0 degrees.
(b) To find the speed of each ball after the collision, we need to use the law of conservation of kinetic energy:
(1/2) * m1 * v1i^2 + (1/2) * m2 * v2i^2 = (1/2) * m1 * v1f^2 + (1/2) * m2 * v2f^2
Substituting the given values:
(1/2) * m1 * (4.0 m/s)^2 + 0 = (1/2) * m1 * v1f^2 + (1/2) * m2 * 0^2
Simplifying this equation, we get:
8.0 m^2/s^2 = (1/2) * m1 * v1f^2
Dividing both sides by (1/2) * m1, we get:
16 m^2/s^2 = v1f^2
Taking the square root of both sides to solve for v1f, we get:
v1f = √16 m/s = 4.0 m/s
For the target ball, we know its final velocity, v2f, will be 0 m/s.
Therefore, the speed of each ball after the collision is:
cue ball: 4.0 m/s
target ball: 0 m/s