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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:

m

So:

v

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.

## 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]m

_{A1}v_{A1}+ m_{B1}v_{B1}= m_{A2}v_{A2}+ m_{B2}v_{B2}So:

v

_{B2}= (m_{A1}v_{A1}+ m_{B1}v_{B1}- m_{A2}v_{A2})/m_{B2}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.

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