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

In summary, the ball bounces twice and the distance between the first and second bounce is the distance between the point of projection and the point of first bounce.
  • #1
j92
2
0

Homework Statement


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


Homework 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.


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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  • #2
how about staying in horizontal/vertical coordinate system - the motion of the ball will be parabolic, the place where it first bounces can be detrimned by finding where a line (the sloping surface) and the parabola intersect.
 
  • #3
Thanks this helped a lot!
 

1. How does the slope of the ramp affect where the ball bounces first?

The steeper the slope of the ramp, the further up the ramp the ball will bounce first.

2. Does the material of the ball impact where it bounces first?

Yes, the material of the ball can impact where it bounces first. A more elastic ball will bounce higher and further up the slope compared to a less elastic ball.

3. What other factors besides slope and ball material can influence where the ball bounces first?

The speed and angle at which the ball is thrown, as well as any external forces such as wind, can also impact where the ball bounces first.

4. Is there a mathematical formula to determine where the ball will bounce first on a slope?

Yes, there is a formula that takes into account the slope of the ramp, the elasticity of the ball, and the initial velocity and angle of the ball to predict where it will bounce first. This is known as the Laws of Motion and can be found in physics textbooks.

5. Can the ball bounce first in multiple places on the slope?

Yes, depending on the initial conditions and the specific characteristics of the ball and slope, the ball may bounce first in multiple places on the slope. This is known as multiple rebounds and can be observed in real-world scenarios such as a basketball bouncing off the backboard before going into the hoop.

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