How Is Average Force Calculated When Bullets Hit Different Surfaces?

[brad sue]

Hi,
Please can I have some help with this problem:

A machine gun in automatic mode fires 20g bullets with v_bullet=300m/s at 60 bullets/second.
a- If the bullets enters a thick wooden wall, what is the average force exerted against the wall?
b- If the bullets hit a steel wall and rebound elastically,what is the average force on the wall?

What i did is
for a:
I computed the impulse: I= p_f-p_i= m*v-( 0)=6 N.s
to calculate the average force: I/(time for one bullet to be fired)= 6/(1/60)=360N

Am I right?
However I have no idea for the second question??
B.

 

[brad sue]

Hi,
Please can I have some help with this problem:
A machine gun in automatic mode fires 20g bullets with v_bullet=300m/s at 60 bullets/second.
a- If the bullets enters a thick wooden wall, what is the average force exerted against the wall?
b- If the bullets hit a steel wall and rebound elastically,what is the average force on the wall?
What i did is
for a:
I computed the impulse: I= p_f-p_i= m*v-( 0)=6 N.s
to calculate the average force: I/(time for one bullet to be fired)= 6/(1/60)=360N
Am I right?
However I have no idea for the second question??
B.

please can someone help me with this problem?

 

[Doc Al]

What i did is
for a:
I computed the impulse: I= p_f-p_i= m*v-( 0)=6 N.s
to calculate the average force: I/(time for one bullet to be fired)= 6/(1/60)=360N
Am I right?

Yes.

However I have no idea for the second question??

It's the same idea. The only difference is the change in momentum. (The final velocity of the bullet is not zero.)

 

[finchie_88]

I think that for the second question, the average force should be twice as much, cos the collision is modeled as completely elastic, so you have the following maths...
FT = m(u-v) // m = 0.02 kg, u = 300 m/s, v = -300 m/s ...
Therefore, the answer to the second question is twice your answer to the first part (I think) someone will correct me if I'm wrong, so if I am wrong, then it doesn't matter