Speed of two bodies that sticks together after collision

Click For Summary
SUMMARY

The problem involves two railway trucks with masses m and 3m colliding and sticking together. The initial speeds are 2v and v, respectively. Using the conservation of momentum, the correct calculation yields a final speed of v/4 after the collision. The initial conclusion of 4/v was incorrect, and the answer provided in the book, 5v/4, is also incorrect based on the given conditions.

PREREQUISITES
  • Understanding of conservation of momentum principles
  • Basic knowledge of mass and velocity in physics
  • Ability to solve algebraic equations
  • Familiarity with the concept of inelastic collisions
NEXT STEPS
  • Study inelastic collision problems in physics
  • Learn about momentum conservation in multi-body systems
  • Review algebraic manipulation techniques for solving equations
  • Explore examples of real-world applications of momentum conservation
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and collision theory, as well as educators looking for examples of momentum conservation in action.

coconut62
Messages
161
Reaction score
1

Homework Statement



Two railway trucks of masses m and 3m move towards each other in opposite directions with speeds 2v and v respectively. These trucks collide and stick together after the collision. What is the speed of the trucks after the collision?

Homework Equations



Conservation of momentum: m1v1 + m2v2 = (m1+m2)Vtotal


The Attempt at a Solution



m(2v) + 3m(-v) = (m+3m)V

V= -4/v
∴ speed = 4/v

Answer given: 5v/4

Apparently they added the 2mv and 3mv together for the initial momentum. Is the book wrong or am I wrong?
 
Physics news on Phys.org
coconut62 said:

Homework Statement



Two railway trucks of masses m and 3m move towards each other in opposite directions with speeds 2v and v respectively. These trucks collide and stick together after the collision. What is the speed of the trucks after the collision?

Homework Equations



Conservation of momentum: m1v1 + m2v2 = (m1+m2)Vtotal


The Attempt at a Solution



m(2v) + 3m(-v) = (m+3m)V

Correct so far.

coconut62 said:
V= -4/v
∴ speed = 4/v

Wrong, do it again.

coconut62 said:
Answer given: 5v/4

Apparently they added the 2mv and 3mv together for the initial momentum. Is the book wrong or am I wrong?

The book is wrong.


ehild
 
From " m(2v) + 3m(-v) = (m+3m)V "

2mv -3mv = 4mV

-mv = 4mV

V= -v/4

Speed does not have a direction, so answer is v/4. Why wrong?
 
You wrote speed=4/v in the original post:biggrin:

ehild
 
coconut62 said:
Answer given: 5v/4.
That would be the right answer if m had speed v and 3m had speed 2v. Are you sure you quoted the question correctly?
 
Yes. I quoted it correctly.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
1K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
335
Views
16K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 29 ·
Replies
29
Views
2K