Is the Book Right? Examining Conservation of Momentum

Click For Summary
SUMMARY

The discussion centers on the application of the conservation of momentum in a physics problem involving a carriage and a student. The proposed solution calculates the speed of the carriage after the student jumps out, yielding the equation u = mv/(M+3m). The original book reference, which states u = mv/(M+2m), is challenged for potentially overlooking relative velocity. Participants emphasize the importance of proper bracket usage in equations and correct momentum balance, clarifying that the ticket collector should not be included on both sides of the equation.

PREREQUISITES
  • Understanding of conservation of momentum principles
  • Familiarity with basic algebra and equation manipulation
  • Knowledge of relative velocity concepts
  • Ability to interpret physics problems involving multiple bodies
NEXT STEPS
  • Review conservation of momentum in multi-body systems
  • Study the impact of relative velocity in physics problems
  • Learn proper equation formatting and the use of brackets in algebra
  • Examine examples from "200 Puzzling Physics Problems" for similar scenarios
USEFUL FOR

Students studying physics, educators teaching momentum concepts, and anyone interested in solving complex physics problems involving multiple objects and relative motion.

phantomvommand
Messages
287
Reaction score
39
Homework Statement
This is a problem from 200 Puzzling Physics Problems.
A student and ticket collector of mass m are in a stationary railway carriage of mass M. The collector realises the student has no ticket, so the student runs towards the end of the carriage, and the collector chases after him with speed v relative to the carriage. The student stops at the end of the carriage and jumps out. Find the velocity of the carriage when the collector stops at the open end of the carriage and watches the student escape.
Relevant Equations
Conservation of Momentum
My proposed solution:

When the student stops at the end, suppose the carriage is moving at speed u.
0 = (M+2m)u - m(v - u)
==> u = mv/ M+3m

After jumping out, the total momentum of the Carriage + collector system is 0 - mu = -m^2v/ M+3m.

By conservation of momentum for the Carriage + collector system, at the time when the collector stops at the open end:

-m^2v/ M+3m = 0 + (M+m)V, where V is the final velocity of the carriage + collector.

V = -m^2v/ (M+3m)(M+m)

The book (200 Puzzling Phys Problems) writes that the speed of the carriage when the student has reached the end and stopped (but not yet jumped) is u = mv/M+2m.

I suppose this is from the Conserve momentum equation: (M+2m)u = mv. Doesn't this forget the fact that v is a relative velocity?

Is the book right, and why am I wrong?

Thank you!
 
Physics news on Phys.org
phantomvommand said:
When the student stops at the end, suppose the carriage is moving at speed u.
0 = (M+2m)u - m(v - u)
==> u = mv/ M+3m
1) your lack of using brackets will cause errors -- if not now, then later
2) Your momentum balance should not have the ticket collector on both sides, but: ticket collector on one side and carriage + student on the other.

##\ ##
 
  • Like
Likes   Reactions: phantomvommand
BvU said:
1) your lack of using brackets will cause errors -- if not now, then later
2) Your momentum balance should not have the ticket collector on both sides, but: ticket collector on one side and carriage + student on the other.

##\ ##
I see my mistake now, thank you very much!
 
  • Like
Likes   Reactions: BvU

Similar threads

  • · Replies 17 ·
Replies
17
Views
963
  • · Replies 27 ·
Replies
27
Views
1K
Replies
21
Views
2K
  • · Replies 4 ·
Replies
4
Views
907
Replies
7
Views
2K
Replies
28
Views
2K
Replies
7
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
10
Views
2K