A cue stick has a mass of 0.5 kg. The cue stick hits the ball with the mass of 0.2 kg at the velocity of 2.5 m/s. What is the velocity of the ball after it is hit?
A. 6.3 m/s
B. 8.3 m/s
C. 2.5 m/s
D. 3.6 m/s
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is calculated by multiplying its mass by its velocity.
Before the collision, the momentum of the cue stick is calculated as:
Momentum of cue stick = mass of cue stick * velocity of cue stick
= 0.5 kg * 0 m/s (since the cue stick is at rest before hitting the ball)
= 0 kg·m/s
The momentum of the ball before the collision is calculated as:
Momentum of ball = mass of ball * velocity of ball
= 0.2 kg * 2.5 m/s
= 0.5 kg·m/s
Total momentum before the collision = momentum of cue stick + momentum of ball
= 0 kg·m/s + 0.5 kg·m/s
= 0.5 kg·m/s
After the collision, the momentum of the cue stick is still 0 kg·m/s since it is at rest after hitting the ball.
Let's assume the velocity of the ball after the collision is v.
Therefore, the momentum of the ball after the collision is calculated as:
Momentum of ball = mass of ball * velocity of ball after the collision
= 0.2 kg * v
Total momentum after the collision = momentum of cue stick after the collision + momentum of ball after the collision
= 0 kg·m/s + 0.2 kg * v
= 0.2 kg * v
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
0.5 kg·m/s = 0.2 kg * v
Simplifying the equation:
v = (0.5 kg·m/s) / 0.2 kg
= 2.5 m/s
Therefore, the velocity of the ball after it is hit is 2.5 m/s.
Answer: C. 2.5 m/s