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Conservation of Momentum Help

  • Thread starter JamesRV
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  • #1
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We just started learning Conservation of Momentum, and I feel completely lost. This is the following question I was asked:

Homework Statement


A 2.00-kg object moving at 6.00 m/s collides with a 4.00-kg object that is initially at rest. After the collision, the 2.00-kg object moves backward at 1.00 m/s. This is a one-dimensional collision, and no external forces are acting.

a) Calculate the velocity of the 4.00-kg object after the collision.

My attempt was to simply use the Momentum equation:

Homework Equations

& The attempt at a solution[/B]

mA1vA1 + mB1vB1 = mA2vA2 + mB2vB2

So:
vB2 = (mA1vA1 + mB1vB1 - mA2vA2)/mB2

Resulting in the velocity to be 3.5m/s.

I was just looking for someone to confirm that answer, because I am unsure if I did it right.

In question b) it asks:
Calculate the velocity of the centre of mass of the two objects (i) before the collision and (ii) after the collision.
What does that mean?

Thank you for helping.
 
Last edited:

Answers and Replies

  • #2
1,540
134
We just started learning Conservation of Momentum, and I feel completely lost. This is the following question I was asked:

A 2.00-kg object moving at 6.00 m/s collides with a 4.00-kg object that is initially at rest. After the collision, the 2.00-kg object moves backward at 1.00 m/s. This is a one-dimensional collision, and no external forces are acting.

a) Calculate the velocity of the 4.00-kg object after the collision.

My attempt was to simply use the Momentum equation:

mA1vA1 + mB1vB1 = mA2vA2 + mB2vB2

So:
vB2 = (mA1vA1 + mB1vB1 - mA2vA2)/mB2

Resulting in the velocity to be 3.5m/s.

I was just looking for someone to confirm that answer, because I am unsure if I did it right.

In question b) it asks:
Calculate the velocity of the centre of mass of the two objects (i) before the collision and (ii) after the collision.
What does that mean?

Thank you for helping.
Hi JamesRV

Welcome to PF!!!

The result of part a) looks right .

For part b) use Vcm = (m1v1+m2v2)/(m1+m2) .

Apply this before and after collision .What do you get ?

Please take care of the signs .
 
Last edited:
  • #3
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Thank you Tanya!

I used the equation you gave me, it makes logic sense taking the total momentum and dividing by the mass.
I got 2m/s for both cases, which seems right!
 
  • #4
1,540
134
Thank you Tanya!

I used the equation you gave me, it makes logic sense taking the total momentum and dividing by the mass.
I got 2m/s for both cases, which seems right!
Yes..Well done :thumbs:

So what does this tell you about collision ?
 
  • #5
4
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Yes..Well done :thumbs:

So what does this tell you about collision ?
Doesn't it just mean that the momentum is conserved? I thought momentum was always conserved, is it not?
 
  • #6
1,540
134
Doesn't it just mean that the momentum is conserved? I thought momentum was always conserved, is it not?
Yes... And the velocity of Center of Mass remains unaffected in a collision.
 
  • #7
4
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Okay, thanks again!
 

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