# Two blocks collide (Momentum-Energy conservation)

• Arman777
In summary, Block 1 of mass 3.0 kg is sliding across a flooe with speed ##v_1=2.00 \frac m s## when it makes a head-on,one dimensional,elastic collision with initially stationary block 2 of mass 2.0 kg.The coefficient of kinetic friction between the blocks and the floor is ##μ_k=0,30##.The speeds of (a) block 1 and 2 just after the collision.Also find (c) their final separation after friction has stopped them and (d) the energy lost to thermal energy because of the friction.
Arman777
Gold Member

## Homework Statement

[/B]Block 1 of mass 3.0 kg is sliding across a flooe with speed ##v_1=2.00 \frac m s## when it makes a head-on,one dimensional,elastic collision with initially stationary block 2 of mass 2.0 kg.The coefficient of kinetic friction between the blocks and the floor is ##μ_k=0,30##
Find the speeds of (a) block 1 and 2 just after the collision.Also find (c) their final separation after friction has stopped them and (d) the energy lost to thermal energy because of the friction.

## Homework Equations

Energy-Momnetum conservation Equations

## The Attempt at a Solution

##Δ\vec p =\vec p_f-\vec p_i=F_{external}Δt##
So If we assume Δt is too small we can say ##Δ\vec p=0## so from that

##m_1v_1=m_1v'_1+m_2v'_2##
##6kg\frac m s=3kg v'_1+2kgv'_2##

And we know there's friction so there must be energy lost.From the Energy conservation
##\frac 1 2m_2(v'_2)^2+\frac 1 2m_1(v'_1)^2-\frac 1 2m_1(v_1)^2=W_{friction}=F_fΔx##

I don't know how to calculate Δx here.Also there's friction between two blocks..Where I should add it in the energy conservation equation ? And How can I calculate it

Last edited:
You have correctly assumed Δt is negligible during the collision. Would that not be true of Δx too?

Arman777 said:
Also there's friction between two blocks.
The problem stated, "The coefficient of kinetic friction between the blocks and the floor is μ_k=0,30." I think what that meant was that there is friction between each block and the floor - not necessarily between the two blocks. The collision is elastic, which is really all you need to know about the interaction of the two blocks.

To me it makes sense to find the velocities immediately after impact. Once the velocities are known, it should be straightforward to solve the remainder of the problem (distance traveled, etc.).

Arman777
haruspex said:
You have correctly assumed Δt is negligible during the collision. Would that not be true of Δx too?

Maybe it took some distance to get there...I mean can we exactly say ##Δx=0## ?

TomHart said:
The problem stated, "The coefficient of kinetic friction between the blocks and the floor is μ_k=0,30." I think what that meant was that there is friction between each block and the floor - not necessarily between the two blocks. The collision is elastic, which is really all you need to know about the interaction of the two blocks.

To me it makes sense to find the velocities immediately after impact. Once the velocities are known, it should be straightforward to solve the remainder of the problem (distance traveled, etc.).

I see you are right about that thefriction betwwen blocks is kinda awkward idea for this quesiton.

Ok I ll solve as ##Δx=0## let's see

Arman777 said:
Maybe it took some distance to get there...I mean can we exactly say Δx=0?
Δx=vΔt. If v is moderate and Δt infinitesimal then Δx is infinitesimal.

haruspex said:
Δx=vΔt. If v is moderate and Δt infinitesimal then Δx is infinitesimal.

but ##Δt## in there is the impact time isn't it ?
and ##Δx## is the distance traveled by object ?

I was thinking like this If there were ##5m## between A and B and they were collide the ##Δt## could be zero or we can assume its zero. But ##Δx=5m## cause it took ##5m## to come there and it lost some kinetic energy ?

By the way,I found the velocities correctly I ll try to solve other parts

Arman777 said:
and Δx is the distance traveled by object ?
The energy equation you posted that involved Δx appeared to be relating energy just before collision to energy just after. Therefore I assumed your Δx represented the distance moved during the collision.
You do need to find the two velocities immediately after the collision.

Arman777
haruspex said:
The energy equation you posted that involved Δx appeared to be relating energy just before collision to energy just after. Therefore I assumed your Δx represented the distance moved during the collision.
You do need to find the two velocities immediately after the collision.

Oh I see now.I never looked that way...

And I solved the question(all parts) Thanks for help

## What is momentum?

Momentum is a measure of an object's motion. It is calculated by multiplying an object's mass by its velocity.

## What is energy conservation?

Energy conservation is the principle that states energy cannot be created or destroyed, it can only be transferred or converted from one form to another.

## How is momentum conserved in a collision?

In a collision, the total momentum of the system (all objects involved) remains constant before and after the collision. This means that the total momentum before the collision is equal to the total momentum after the collision.

## How is energy conserved in a collision?

In a collision, the total energy of the system (all objects involved) remains constant before and after the collision. This means that the total energy before the collision is equal to the total energy after the collision.

## What factors can affect the conservation of momentum and energy in a collision?

The conservation of momentum and energy in a collision can be affected by external forces, such as friction or air resistance, and the elasticity of the objects involved.

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