A cue stick hits a cue ball with an average force of 22 N for a duration of 0.023 s. If the mass of the ball is 0.18 kg, how fast is it moving after being struck?
2006-10-15 11:18:09 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Use F=ma to determine what the acceleration of the cue is.
Then use v = at to determine what the velocity becomes.
I get 2.8 m/s. (remember significant figures)
Aloha
2006-10-15 11:22:04 · answer #1 · answered by Anonymous · 2⤊ 0⤋
This is a question of a change in momentum. Momentum is changed when an "impulse" is applied. In your case, an impulse is:
(average force applied)*(time the force is applied)
Thus, your impulse is:
(22 Newtons)*(0.023 seconds) = 0.506 kg*m/s
Your cue ball is 0.18 kg. Its new momentum is given by:
(mass)*(velocity)
Since it started at 0 velocity, its initial momentum is 0 kg*m/s. It's new momentum is thus 0.506 kg*m/s. So divide 0.506 kg*m/s by its mass to get its new velocity:
new velocity = impulse/mass = ( 0.506 kg*m/s )/(0.18 kg) = 2.81 m/s
The cue ball will move 2.81 m/s after it is struck.
2006-10-15 18:44:22 · answer #2 · answered by Ted 4 · 0⤊ 0⤋
Force = mass * acceleration = mass * (change in velocity/change in time)
22 N = (0.18 kg) * (v/0.023 s)
Solve for v
v = 2.811 m/s
2006-10-15 18:22:21 · answer #3 · answered by TM 3 · 1⤊ 0⤋