Question

I am asked to show if momentum is conserved in this scenario. I'll post the picture and my calculations.

Answer 1

So I think the momentum before the collision is 0.35 kgm/s and it is the same after the collision because...

0.5 * -0.35 = -0.175 kgm/s for ball 1
1.5 * 0.35 = 0.525 kgm/s for ball 2
Adding the two we get 0.35 kgm/s

Answer 2

momentum is **always** conserved !

are you just looking for help in running the numbers ?!

Answer 3

Yes, I need the calculations to show that momentum is conserved.

Answer 4

well in the \(v_y\) columns nothing is happening -- that's the up and down direction -- so i reckon we can ignore that bit of information. yes?

Answer 5

Yes, there is another section on the assignment for nonlinear collisions, so the Vy part is changing only for that. We can ignore it here.

Answer 6

in the x direction -- left to right --

we start with linear momentum is:

\(0.5 \cdot 0.7 = 0.35\)

after the collision we have, according to my calculator, this linear momentum:

\(0.5\cdot(-0.088) + 1.5\cdot(0.263) = 0.35 \color{red}{05}\)

that's a rounding error, right!

Answer 7

Must be a rounding error. I messed up my calculations by not using the Vx they provided after the collision.

Answer 8

I'm guessing that the computer rounded off a decimal to -0.088 and 0.263.

Answer 9

indeed

\(0.0\dot 8 = \frac{8}{90} = \frac{4}{45}\)

same will go for the 0.263, prolly rounded too

it is in fact \(\frac{11}{54}\) !!

Answer 10

just remember, if momentum is not conserved --- Newton's 3rd Law, his master-stroke --- then your numbers are wrong 😋

but yeah, rounding, and website/whatever data-entry, must be annoying :(

Answer 11

Haha...definitely