Momentum: A bullet passing through a wooden block on a frictionless surface

Click For Summary

Homework Help Overview

The problem involves a wooden block on a frictionless surface being impacted by a projectile. The scenario includes calculating the final velocities of both the block and the projectile after the impact, considering the forces involved and the conservation of momentum.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply kinetic energy and momentum conservation principles to solve for the final velocities. Some participants question the correctness of the energy balance and momentum equations used. Others suggest verifying the setup and equations to ensure they align with the physical scenario described.

Discussion Status

Participants are actively discussing the equations needed to approach the problem. Some guidance has been provided regarding the correct equations to use, but there is still uncertainty about how to proceed with solving the equations. The discussion reflects a mix of interpretations and attempts to clarify the problem setup.

Contextual Notes

There are indications of confusion regarding the application of energy and momentum conservation principles, as well as the complexity of the equations involved. The original poster expresses difficulty in managing the equations derived from the conservation laws.

mmoadi
Messages
149
Reaction score
0

Homework Statement



A wooden block with a length of 10 cm and a mass of 1 kg lies on an icy plane. A projectile with mass of 2 g hits the wooden block with velocity of 300 m/s and breaks through its center of gravity. How much are the final velocities of the wooden block and the projectile? While moving through the wooden block, the projectile worked on it with a force of 500 N. This is a frictionless system.

Homework Equations



KE= ½ mv²
p=mv

The Attempt at a Solution



KE= F*d=50 J
KE= ½ m(1)v(final)² → v(1-final)²= 2KE/ m(1)
v(1-final)= 223.6 m/s

Conservation of the momentum:
m(1)v(1-initial) + m(2) v(2-initial) = m(1)v(1-final) + m(2) v(2-final)
v(2-final)= [m(1)v(1-initial) + m(2) v(2-initial) - m(1)v(1-final)] / m(2)
v(2-final)= 0.15 m/s

Is my approach to the problem and calculations correct?
Thank you for helping!
 
Physics news on Phys.org
Your energy balance equation is incorrect. The initial kinetic energy of the bullet KE0, which you can calculate is a certain number of Joules. That number of Joules is divided into three parts

1. Work done by bullet on block (F*d)
2. Final kinetic energy of block.
3. Final kinetic energy of bullet.

Your momentum conservation equation is also incorrect.

Pbefore=Momentum of bullet only
Pafter=Momentum of bullet + momentum of block

"Before" means before the bullet hits the block; "after" means after the bullet has made it through.
 
So, if I write it like this:

½ m(1)v(1-initial)²= ½ m(1)v(1-final)² + ½m(2)v(2-final)²+F*d

And

m(1)v(1-initial)= m(1)v(1-final) + m(2)v(2-final)

I have the right equations to work with?

Thank you for helping!
 
Last edited:
Yes, these are the correct equations to start from.
 
OK, but how do I continue.:confused:
I tried to express v(2-final) with the components of the second formula (conservation of the momentum) and insert it into the first formula (KE) but it just got very complicated and weird.:redface:

Can you give me another hint, please?
 
mmoadi said:
OK, but how do I continue.:confused:
I tried to express v(2-final) with the components of the second formula (conservation of the momentum) and insert it into the first formula (KE) but it just got very complicated and weird.:redface:

Can you give me another hint, please?

Weird or no weird, that's exactly what you have to do. There are no other hints. You have to solve a system of two equations and two unknowns and you described the correct way to solve it.
 
Thank you for helping! :smile:
Back to solving my weirdness of equation!:biggrin:
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
949
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
27
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K