Conservation of linear momentum problem

Click For Summary

Homework Help Overview

The problem involves the conservation of linear momentum in the context of an explosion where a particle of mass 5m splits into two pieces of masses 2m and 3m. The lighter piece moves with a speed of 5v relative to the heavier piece, and the goal is to determine the actual speed of the lighter piece.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum and question the interpretation of relative speed in the context of the problem. There are attempts to set up equations based on the initial and final momentum, and some participants express confusion regarding the direction of motion and the meaning of "relative speed."

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and attempting to clarify the relationships between the velocities of the two pieces after the explosion. Some guidance has been offered regarding the setup of equations, but no consensus has been reached on the correct approach.

Contextual Notes

Participants note confusion around the term "relative speed" and its implications for the velocities of the two pieces post-explosion. There is also mention of an impending test, which may add pressure to resolve the problem quickly.

risingabove
Messages
20
Reaction score
0

Homework Statement


A particle of mass 5m moving with speed v explodes and spilts into two pieces with masses of 2m and 3m. The lighter piece continues to move in the original direction with speed 5v relative to the heavier piece. what is the actual speed of the lighter piece?



Homework Equations


using law of conservation of linear momentum
initial moment = final moment
mass (m) x Velocity (v) = mass (m) x Velocity (v)


The Attempt at a Solution



using the formula...

5m (v) = 2m(v) - 3m (5v) ...) i used the minus (-) sign because the heavier piece went a different direction

5mv = 2mv -15mv

20mv = 2mv

10v = v

but my answer was wrong...

PLEASE HELP EXPLAIN WHY AND WHAT IS IT IM MISSING TO DO>>>

FIRST CLASS TEST IS TOMORROW!
 
Physics news on Phys.org
5m (v) = 2m(v) - 3m (5v)

That's not correct. You appear to be assuming that after the explosion the large bit is moving five times faster than the small bit. That's not what the problem states.

i used the minus (-) sign because the heavier piece went a different direction

I wouldn't recommend that. I would just define

Vi is the initial velocity
Vsmall as the new velocity of the small bit
Vlarge as the new velocity of the large bit

So with that in mind it should be something like

5mVi = 2mVsmall + 3mVlarge

I'll let you figure out the other equation to write.

Sorry for the edits to my reply.
 
ok soo would i know say

5mVi - 3mVlarge = 2mVsmall

Ok i think I am confusing myself but i do get and understand what you were saying...i think the "relative" word in the problem is what confusing me...

what do they mean when they say "...with a speed 5v relative..."
 
Pleas Helppp...i really want to understand this
 
Im still stuck on this question can someone give any insights..? PLEASE...
 
risingabove said:
ok soo would i know say

5mVi - 3mVlarge = 2mVsmall
Yes
Ok i think I am confusing myself but i do get and understand what you were saying...i think the "relative" word in the problem is what confusing me...

what do they mean when they say "...with a speed 5v relative..."
it's how much faster it's moving. Vsmall = Vlarge + 5v.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
952
  • · Replies 2 ·
Replies
2
Views
1K
Replies
335
Views
16K
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
23
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K