Conservation of Momentum with Friction

Click For Summary
SUMMARY

The discussion focuses on the conservation of momentum in a collision involving a bullet and a block on a rough surface. A bullet with a mass of 6.32 g is fired into a block of mass 2.30 kg, which is initially at rest, and they slide together until coming to rest due to friction with a coefficient of 0.28. The participant attempted to apply the momentum equation and kinetic energy principles but encountered issues with the calculations, particularly in accounting for friction. The key takeaway is that the distance traveled while coming to a stop is essential for accurately calculating the effects of friction on kinetic energy.

PREREQUISITES
  • Understanding of conservation of momentum principles
  • Knowledge of kinetic energy equations
  • Familiarity with friction coefficients and their impact on motion
  • Basic algebra for solving equations
NEXT STEPS
  • Study the concept of inelastic collisions and momentum conservation
  • Learn how to calculate work done by friction and its effect on kinetic energy
  • Explore the relationship between distance, force, and energy in motion
  • Investigate the equations of motion under constant acceleration
USEFUL FOR

Physics students, educators, and anyone interested in understanding the principles of momentum and energy in collision scenarios, particularly in the context of frictional forces.

derf709
Messages
1
Reaction score
0

Homework Statement



A bullet of mass m = 6.32 g is fired into a block of mass M = 2.30 kg that is initially at rest on a rough surface. (The coefficient of friction is 0.28 between the block and the surface.) The bullet ends up stuck in the block, and together they slide across the surface and come to rest.

Homework Equations



KE=1/2 mv^2
p= mv

The Attempt at a Solution



(.00632)(vi)=(.00632+2.30)(Vf)

I was thinking I could set the total kinetic energy minus friction equal to zero, and then solve for the velocity, which could be used in the momentum equation. is this on the right track?

1/2(.00632+2.30)v^2-6.33=0
v=2.34 m/s

Plugging in final velocity, I got an initial velocity of 855 m/s. It sounds plausible, but the answer is incorrect. Any suggestions? :)
 
Physics news on Phys.org
Your problem is missing information. You can't subtract friction force in Newtons from KE which is in in Newton-meters (joules). You need to know the distance traveled while coming to a stop.
 

Similar threads

Conservation of momentum (wrecking ball hits a stationary object)
  • · Replies 10 ·
Replies
10
Views
3K
Linear Motion and Linear Momentum
  • · Replies 31 ·
2
Replies
31
Views
4K
A bullet collides perfectly elastically with one end of a rod
  • · Replies 21 ·
Replies
21
Views
3K
State whether the distance x for the block would be greater/less/same
  • · Replies 25 ·
Replies
25
Views
2K
Choosing what consists of a "system" in Newton's laws of motion
  • · Replies 28 ·
Replies
28
Views
2K
Angular momentum for a bullet and a rod attached to a ceiling
  • · Replies 17 ·
Replies
17
Views
2K
Dropping a block on another block which is performing SHM
  • · Replies 17 ·
Replies
17
Views
2K
An exploding cannon shell
  • · Replies 4 ·
Replies
4
Views
911
Angular Momentum- Conserved or not?
  • · Replies 17 ·
Replies
17
Views
1K
Finding the work done by a block
  • · Replies 4 ·
Replies
4
Views
3K