A block moving on the floor and hitting the walls

[Arthur]

Oh ok… so now how do we find the velocity after collision from wall 1? If we are able to find that then I can apply same procedure to collision from wall 2…

[Alisher (Alisher Yerkebayev)]

@Alisher

elastic collision implies that the speed of the block hasn’t changed

but if the final answer gets wrong, then the assumption that the collision is elastic is incorrect

[Arthur]

All collision are assumed elastic

[Arthur]

So that means the velocity will be V0 after colliding from wall 1 and then the block collides with wall 2, so what will we do in second collision to find the final answer?

[Alisher (Alisher Yerkebayev)]

absolutely the same thing

[Arthur]

But we know that collision are assumed elastic so after collision from wall 2 there should not be any change in velocity of block but … The answer given is 6 m/s as the final velocity of block after collision with wall 2.

[Arthur]

Could you explain how to get 6 m/s as the velocity of block after collision with wall 2? I am not able to get that.

[Alisher (Alisher Yerkebayev)]

lol, i have really forgot that we had to calculate the final velocity, sorry.

@Alisher

the perpendicular projection of velocity is v_0^n=v_0\sin37 and apparently after collision it has become v_1^n​=−v_0^n​ (i assume that the collision is elastic, but i’m not sure whether it is allowed or not). Thus the change of momentum is

\int N dt = m(v_0^n-v_1^n) = 2mv_0\sin37.

use the very first assumption, with this you can get the desired answer, i’ve already checked it and got 5.98 m/s

[Arthur]

Oh ok thank you! I am trying again… I’ll tell you if I get the answer correctly…

[Arthur]

So the block will strike the wall 2 at angle of 34 degrees (measured fr

om the wall 2). Is this diagram correct?

[Alisher (Alisher Yerkebayev)]

i had 44.8\degree

[Arthur]

Could you send your diagram for more clarity?

[Alisher (Alisher Yerkebayev)]

just in place where you wrote 34\degree, i have 44.8\degree

[Arthur]

But how did you get 44.8 ? I am unable to understand. Could you send solution for second collision with wall 2 please?

[Arthur]

I know I am troubling you a lot but actually I am not that strong in physics so I need a lot of help, hope you dont mind😊

[Alisher (Alisher Yerkebayev)]

the projection of velocity v_\tau along wall 1 equals v_0(\cos37-2\mu\sin37) (try to get the same result), so \tan\beta = v_\tau/v_0\sin37 = \cot37 -2\mu = 0.9937

@Arthur

I know I am troubling you a lot

nevermind, it’s ok

[Arthur]

So β is the reflection angle of block from wall 1?

[Alisher (Alisher Yerkebayev)]

yes, but it is relative to the normal line because it is essentialy the same angle of incidence (like 37\degree, but with respect to wall 2)

[Arthur]

Oh ok…

[Arthur]

I am applying same procedure to wall 2 as well.