# Elastic collisions help

Summarising my working, I said that after the first collision the velocity component parallel to the cushion is unchanged so remains as u cos(20) and the perpendicular component u sin(20) becomes (1/2)u sin(20).

Then in the second collision the direction of the components switch around and so (1/2)u sin(20) remains the same and u cos(20) becomes (2/5)u cos(20).

Then using Pythagoras, the final speed^2 = (1/2)^2 u^2 + (2/5)^2 u^2 = 41/100 u^2 and so the kinetic energy has decreased by 59%.

The answer given in the textbook is 83% and the textbook's method is much more complicated than mine so I'm thinking that my understanding is wrong somewhere.
(edited 1 year ago)
Original post by 0-)

Summarising my working, I said that after the first collision the velocity component parallel to the cushion is unchanged so remains as u cos(20) and the perpendicular component u sin(20) becomes (1/2)u sin(20).

Then in the second collision the direction of the components switch around and so (1/2)u sin(20) remains the same and u cos(20) becomes (2/5)u cos(20).

Then using Pythagoras, the final speed^2 = (1/2)^2 u^2 + (2/5)^2 u^2 = 41/100 u^2 and so the kinetic energy has decreased by 59%.

The answer given in the textbook is 83% and the textbook's method is much more complicated than mine so I'm thinking that my understanding is wrong somewhere.

It sounds like youre assuming the angle after the first collision is unchanged, but as the parallel/perpedicular components change so will the (complementary) 20 for the second cushion. Also, you seem to ignore the trig part of the components when doing pythagoras at the end?

(edited 1 year ago)
Original post by mqb2766
Also, you seem to ignore the trig part of the components when doing pythagoras at the end?

Thanks I found my mistake.

(1/2)^2 x sin^2(20) x u^2 + (2/5)^2 cos^2(20) x u^2

and thought the trig disappears due to sin^2(20) + cos^2(20) = 1 but that's clearly wrong. If I use my calculator for the above I get 0.1705...u^2 which gives 83% like the textbook. I thought it was odd that my answer didn't depend on the initial angle!

Here's my diagram:

I didn't really think much about the angles and just thought about how the components change. Can you please give more details about the incorrect assumption I made?

And here's the textbook solution:

Here's the textbook solution:
Original post by 0-)
Thanks I found my mistake.

(1/2)^2 x sin^2(20) x u^2 + (2/5)^2 cos^2(20) x u^2

and thought the trig disappears due to sin^2(20) + cos^2(20) = 1 but that's clearly wrong. If I use my calculator for the above I get 0.1705...u^2 which gives 83% like the textbook. I thought it was odd that my answer didn't depend on the initial angle!

Here's my diagram:

I didn't really think much about the angles and just thought about how the components change. Can you please give more details about the incorrect assumption I made?

And here's the textbook solution:

Here's the textbook solution:

For the sketch, Id put on the resolved components before the first collision, after the first and after the second collision as well as the angles and magnitudes. Obviously you have to work some of these out so just use letters as appropriate. In a sense if youre working perp/parallel its not necessary to work out the angles for this question part, but indicate that on your sketch.

Your solution was simpler than the worked solution as by splitting into perp/parallel, and applying restitution to each dimension as appropriate for the two (perpendicualr) cushions. As you say, youve sorted the pythagoras problem.

The textbook solution seems to work in speed/angle coordinates for each part which is simply unnecessary in this case as the cushions are pependicular.
(edited 1 year ago)