What is the peak force between two pool balls in a pool shot?

[fk378]

Homework Statement

What is the peak force between two pool balls in a pool shot? Estimate the momentum transferred to the object ball in a hard pool shot. Estimate (or look up) masses and velocities. For how long are two pool balls in contact? The time can be approximated by the length of time it takes a sound wave to cross a pool ball. Put it all together and compare it to the force of gravity or hte normal force from the table on each ball.

Homework Equations

p=mv
F=dp/dt

The Attempt at a Solution

Estimations:
mass of cue ball- .16 kg
velocity of cue ball- .7 m/s
-->p=.229 kg m/s
speed of sound- 340.29 m/s
length of cue ball- (2.25 in)(.0254 m/in)=.05715 m
-->speed of sound across a cue ball=1.68E-4 s
--> F=dp/dt=(.229 kg m/s)/(1.68E-4 s)=1360 N
and
F_g=mg=(.16 kg)(9.81 m/s^2)=1.57 N


The force that I got is WAY too high, but I don't know what I did wrong.

 

[joeyar]

You have made an error calculating the momentum. The equation is p = mv, but it looks like you calculated m/v.

When I calculated the momentum, I got 0.112 kg m/s, not 0.229 kg m/s

 

[fk378]

Well, now when I calculated the force I get 667 N. Which is still pretty high. And much higher than F_g. Can this be right?

 

[borgwal]

Of course the force is much larger than mg: compare the forces you feel when (i) you gently put a pool ball at rest on your head (=mg), and (ii) the pool ball hits your head at some typical speed and bounces off.