Question

A 1 kg mass traveling at 2 m/s collides head on with a 3 kg mass traveling at 4 m/s. What is the magnitude of the final momentum (in kg m/s) of the 3 kg mass if the one kg mass rebounds in the opposite direction at 2 m/s?

Answer 1

Not sure if I'm doing this right.
1kg * 2 m/s = 2 kg m/s
3kg * 4 m/s = 12 kg m/s
Total momentum initial = Total momentum final
12 kg m/s = 12 kg m/s
So...
1kg * 2 m/s = 2 kg m/s
12kg m/s - 2kg m/s = 10 kg m/s

Is 10 kg m/s the right answer?
Or is that wrong and I need to add the 2kg m/s in that it is moving the opposite direction. HELP.

Answer 2

total momentum of the "system" i.e. the two objects
is
\[ m_1 \cdot v_1 + m_2\cdot v_2\]
based on the question, we know the velocities are opposite directions
so we can let one be negative and the other positive

Answer 3

So I'm thinking now that the initial momentum is -10.

Answer 4

Which would make the final momentum -10?

Answer 5

And then if the 1 kg mass has a momentum of -2 Ns. The other will have a momentum of -8 Ns. Which is a magnitude of 8?

Answer 6

so you can say
1*-2 + 3*4 = -2 +12=10
(which velocity is negative is arbitrary)

Answer 7

Makes sense

Answer 8

That should essentially be the same right?

Answer 9

momentum is conserved, so we can say
1*2 + 3*x = 10
we use +2 because we are going the opposite way of -2)
3x=8

magnitude of the final momentum is | 8| = 8

Answer 10

Sweet awesome. You were very helpful!

Answer 11

if we started with
1*2+3 *-4 = -10
1*-2 + 3x= -10
3x= -8
|-8| i.e. the momentum = 8

Answer 12

Yep. I was thinking they would mathematically end up the same either way. Thanks again!