Momentum and Stream of Pellets

[bob1182006]

Not really homework but doing all problems in the book this one just stumped me...

Homework Statement

A pellet gun finers ten 2.14-g pellets per second with a speed of 483 m/s. The pellets are stopped by a rigid wall
(a) Find the momentum of each pellet.
(b) Calculate the average force exerted by the stream of pellets on the wall.
(c) If each pellet is in contact with the wall for 1.25 ms, what is the average force exerted on the wall by each pellet while in contact? Why is this so different from (b)?

Homework Equations

P=mv
P2-P1=Favg*(t2-t1)

The Attempt at a Solution

$$(a) P=m_pv_p=1.03kg*m/s$$
(b) ok the "stream" screws me up, wouldn't the average force of the stream be the average force of just 1? it's just that it's one after the other etc...
$$F_{avg}=\frac{P_2-P_1}{\Delta t}=\frac{m_P(V_f-V_i)}{\Delta t}$$
I don't know a delta t, final velocity=0?

(c)$$f_{avg}=\frac{m_P(V_f-V_i)}{\Delta t}=\frac{-483*2.14*10^-3}{1.25*10^{-3}}=-826.9N$$
this is force from bullet to wall, wall to bullet = 826.9N

 

[Dick]

The force in c) is only exerted on the wall for a short time. Most of the time the force on the wall is zero. b) is asking for the average force over long periods of time. You could consider one bullet with a delta t of 1/10 sec or ten bullets with a delta t of one second. You'll get the same answer either way.

 

[bob1182006]

I see thanks!
So pretty much for 1s, 10 pellets hit the wall, but they're only in contact for about 1-2 ms, so 10-20ms of the 1s is when the force is applied, and the other time when no force is experienced by the wall brings down the average force to such a low level (-10N).
Wierd H&R, nor my teacher talked about these problems, and they're even numbers so can't look at sol. man.
Time to go try some more ^.^