How Is Average Force Calculated When Bullets Hit Different Surfaces?

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The discussion revolves around calculating the average force exerted by bullets when they hit different surfaces. For a thick wooden wall, the average force calculated is 360 N based on the impulse formula. When bullets hit a steel wall and rebound elastically, the average force is suggested to be double that of the wooden wall, due to the change in momentum being greater in an elastic collision. The calculations involve using the mass and velocity of the bullets to determine the forces. The conversation emphasizes the importance of understanding impulse and momentum in these scenarios.
brad sue
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Hi,
Please can I have some help with this problem:

A machine gun in automatic mode fires 20g bullets with vbullet=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= pf-pi= 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.
 
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brad sue said:
Hi,
Please can I have some help with this problem:
A machine gun in automatic mode fires 20g bullets with vbullet=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= pf-pi= 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?
 
brad sue said:
What i did is
for a:
I computed the impulse: I= pf-pi= 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.)
 
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
 
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