Linear momentum- is my answer correct?

Click For Summary
SUMMARY

The discussion centers on a physics problem involving two blocks with masses of 1 kg and 4 kg colliding on a frictional surface, with a spring attached to the 4 kg block. The calculated maximum compression of the spring after the collision is 0.396 m, based on the conservation of momentum and energy principles. However, there is a discrepancy with a review book that claims the final velocity of the blocks should be 53.6 m/s, which is deemed excessively high by participants. The consensus suggests that the final velocity should be 5.6 m/s, indicating a misunderstanding in energy transfer during the collision.

PREREQUISITES
  • Understanding of conservation of momentum in inelastic and elastic collisions
  • Knowledge of kinetic energy and potential energy relationships
  • Familiarity with Hooke's Law and spring constants
  • Basic algebra for solving equations
NEXT STEPS
  • Review the principles of conservation of momentum in elastic and inelastic collisions
  • Study the relationship between kinetic energy and potential energy in spring systems
  • Learn about Hooke's Law and its applications in collision problems
  • Explore common misconceptions in physics problems involving energy transfer
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding collision dynamics and energy conservation principles.

imatreyu
Messages
79
Reaction score
0

Homework Statement



Two blocks with masses 1 kg and 4 kg, respectively, are moving on a horizontal frictional surface. The 1-kg block has a velocity of 12 m/s, and the 4-kg block is ahead of it, moving at 4m/s. The 4 kg block has a massless spring attached to the end facing the 1-kg block. The spring has a force constant k equal to 1000 N/m.

a. What is the maximum compression of the spring after the collision?
b. What are the final velocities of the blocks after the collision has taken place?

Homework Equations



conservation of momentum (inelastic and elastic)

The Attempt at a Solution



a.
For an instant, the collision is inelastic
mv1i + Mv2i = (m+M) vf
Solve for vf--> vf= 5.6 m/s

All kinetic energy goes into compressing the spring:
.5mv^2=.5kx^2

--> mv^2=kx^2
(m+M)vf^2=1000N/mx^2

Solving for x gets .396 m.

My review book uses the same method as me but says that final velocity should be 53.6 m/s. And thus, x= .05 m. Am I correct or is the book correct? Also, it doesn't matter what units mass is in, right?
 
Last edited:
Physics news on Phys.org
The final velocity of what is supposedly 53.6 m/s? Surely not the 4kg block?
 
imatreyu said:
a.
For an instant, the collision is inelastic
mv1i + Mv2i = (m+M) vf
Solve for vf--> vf= 5.6 m/s

This part seems correct to me, although I wouldn't call it 'for and instant'. It's the total time for the spring to compress. But maybe we're talking semantics here.

imatreyu said:
All kinetic energy goes into compressing the spring:
.5mv^2=.5kx^2

I think you should reconsider this part. If all the kinetic energy goes into the spring as potential energy then both blocks would be at rest. However, both blocks were moving in the same direction and had a collision. By what mechanism would would both blocks come to rest? Even if the 1kg block was brought to a stop by the spring force, the 4 kg block would be pushed from the rear increasing its velocity. I think it would be correct to say that 'some of the kinetic energy goes into compressing the spring.'

imatreyu said:
My review book uses the same method as me but says that final velocity should be 53.6 m/s.

gneill is skeptical of this answer and so am I. I think it's way too high. I got an answer that was less than 10m/s.
 

Similar threads

Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
10
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
4K
Replies
17
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
335
Views
16K
  • · Replies 10 ·
Replies
10
Views
1K