Calculating Force and Thickness of a 8.0g Bullet Passing Through Wood

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving an 8.0g bullet passing through wood, focusing on calculating the force exerted by the block on the bullet, the force exerted by the bullet on the block, and the thickness of the bullet itself. The context includes kinematics and dynamics principles.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the bullet and the block, with one participant expressing uncertainty about how to determine the thickness of the bullet. Another participant suggests that the question might be a typo, proposing that it should refer to the thickness of the wood instead.

Discussion Status

Some participants have provided calculations for the forces involved, while others are exploring the interpretation of the problem. There is no explicit consensus on the thickness question, but there is a productive exchange regarding the calculations and assumptions made.

Contextual Notes

Participants are working under the assumption that the motion of the wood can be ignored, and there is a potential ambiguity regarding the question about the thickness of the bullet versus the wood.

Hollysmoke
Messages
185
Reaction score
0
A 8.0g bullet traveling at 400m/s passes through a heavy block of wood in 4.0x10^-4s, emerging with a velocity of 100m/s. Ignore the motion of the wood.

1) What is the force exerted by the block on the bullet?
2) What is the force exerted by tbe bullet on the block?
3) How thick is the bullet?

I've gotten 1 and 2 but I have no idea how to solve the 3rd one. Can someone please help me out?
 
Physics news on Phys.org
a= -750,000m/s^2
F = 6000N

Just to save time
 
I'm going to take a wild guess and say that the last question is a typo and that it should have read: How thick is the wood? :wink:
 
I really hope so, because the sheet says "bullet".
 
If Icalculate the wood, do I do the following:

v2^2=v1^2 + 2ad
100^2 -400^2 = 2(-750,000)d

d=0.1m
 
Looks good to me.
 
Also, for this question:

The Easter Bunny runs along a straight and narrow path with a constant speed of 25m/s. He passes a sleeping tortoise, which immediately starts to chase the bunny with a constant acceleration of 3x10^-3m/s^2. How long does it take to catch up to the bunny?

I did this:

v=dt
d=vt
d=25m/st

25t= 1/2(3.0x10^-3m/s^2)t^2
t = 16,667 seconds, or 4.63 hours.
 
Looks good. (Just be careful when you are writing up your steps--assuming you need to show your work. Don't write something like "v=dt" when you mean "v=d/t".)
 
Okay. Thanks for confirming my answers
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
14
Views
4K
  • · Replies 4 ·
Replies
4
Views
6K
Replies
3
Views
4K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
1
Views
2K
Replies
3
Views
3K
  • · Replies 28 ·
Replies
28
Views
9K
  • · Replies 4 ·
Replies
4
Views
5K