(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

An elastic ball is projected with speed V from a point O of a smooth plane inclined at an angle theta to the horizontal. The initial direction of projection is up the plane, makes an angle theta + phi with horizontal and lies in a vertical plane containing a line of greatest slope of the plane. (i)Calculate the distance between O and the point of the first bounce.

(ii)If the ball ceases to bounce at the instant it returns to O show that tan(theta)tan(phi)=1-e

2. Relevant equations

The way I tried this for both parts was to make the x axis follow the line of the slope NOT the horizontal which means splitting gravity into components, but this became really confusing.

3. The attempt at a solution

I chose the x axis to be the line of slope and z to be at right angles

to this as the other axis. Using F=ma and F=mg I found a=g and the

integrated to find the position vector r(t) = g(t^2/2)+vo, where vo is

the starting velocity in component form, i.e. vo = kVsin(phi) +

iVcos(phi), where k and i are unit vectors. Then because gravity has

to be split up I used vector g = -igsin(theta)-kgcos(theta). Plugging

both vector g and vo into the r(t) equation and splitting again into

components, I get z(t) = Vtsin(phi)-(t^2/2)gcos(theta) and

x(t)=Vtcos(phi)-g(t^2/2)sin(theta). When the ball hits the slope z(t)

= 0 so using my equation for z(t) and rearranging to get t, I find

t=2Vsin(phi)/gcos(theta). Then I put this t into x(t) and that should

be the distance up the slope. To find the distance from O along the

horizontal I used trig, but it comes out with this horrible looking

equation and I'm pretty sure my error is in the gravity stuff because

the change in axis confuses me massively! Sorry this is really

rambling, I hope it makes sense.

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# An elastic ball bouncing up a slope, where does it bounce first?

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