M4 edexcel, ex 2A question 8

Well I've had a good go at it and can't get it.

8) Two smooth vertical walls stand on a smooth horizontal floor and intersect at an acute angle θ. A small smooth particle is projected along the floor at right angles to one of the walls and away from it. After one impact with each wall the particle is moving parallel to the first wall it struck. Given that the coefficient of restitution between the particle and each wall is e, show that;

(1+2e)tan²θ = e²

Reply 1

Christophicus

anyone?

Reply 2

Christophicus

Please help me.

Reply 3

spoonboy

Widowmaker

Please help me.

The question involves solving 4 symultaneous equations.
I can't remember much of it. To begin with call the angle between the walls θ, then the angle of the first impact is 90 - θ. sin(90 - θ ) = cos θ.

Reply 4

dvs

Suppose that the ball is initially moving with speed u, after it strikes the first wall its speed becomes v, and after it strikes the third wall its speed becomes v'. Then:

(i) First collision
Momentum: u sinθ = v cosy (see attachment to see what angle 'y' is)
Restitution: v siny = eu cosθ
=> tanθ/e = coty
=> tany = e/tanθ

(ii) Second collision
Momentum: v' cosθ = u sinz ('z' in attachment)
Restitution: v' sinθ = eu cosz
=> tanθ = e tanz (2)

With some geometry, you'll see that:
z = 180 - (90 + 180 - (90 - θ + y)) = y - θ

So (2) becomes:
tanθ = e tan(y - &#952 :wink:

= e (tany - tan&#952 :wink:

/(1 + tanytan&#952 :wink:

tanθ + tanytan²θ = etany - etanθ

But from (1), tany=e/tanθ, so:
tanθ + etanθ = e²/tanθ - etanθ
tanθ + 2etanθ = e²/tanθ
(1+2e)tan²θ = e²

pic.PNG5.0KB

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