A cue stick hits a cue ball with an average force of 22.6 N for a duration of 0.0268 s. If the mass of the ball is 0.148 kg, how fast is it moving after being struck?
2007-03-18 11:59:03 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Momentum P = Ft=mv
F-force
t - duration over which the force was applies
V -speed
m- mass of the object
Then v=Ft/m
v=22.6 .0268 / 0.148=4.09 m/s
2007-03-18 12:06:25 · answer #1 · answered by Edward 7 · 1⤊ 0⤋
force=mass*distance/time^2
so you want to divide the force by the mass and multiply by the time for the velocity
I am getting about 4.09 m/s
2007-03-18 19:04:12 · answer #2 · answered by MLBfreek35 5 · 0⤊ 0⤋
impulse = changing momentum
Ft = mv
22.6 * .0268 = .148 *v
v = 4.09m/s
2007-03-18 19:05:25 · answer #3 · answered by 7 · 0⤊ 0⤋
22.6*.0268 = .148*x
x = 4.092 m/s
2007-03-18 19:02:36 · answer #4 · answered by Anonymous · 0⤊ 0⤋