Conservation of linear momentum problem

AI Thread Summary
A particle of mass 5m moving at speed v explodes into two pieces with masses 2m and 3m, where the lighter piece moves at a speed of 5v relative to the heavier piece. The conservation of linear momentum must be applied, leading to the equation 5m(v) = 2m(Vsmall) + 3m(Vlarge). The confusion arises from the term "relative," which indicates that the speed of the lighter piece is 5v faster than the heavier piece, expressed as Vsmall = Vlarge + 5v. Correctly setting up the equations and understanding the relative velocity is crucial to solving the problem. Clarifying these concepts will help in finding the actual speed of the lighter piece.
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top