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Ball bouncing on inclined ramps

  1. Mar 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Last week we did a practical in which we bounced a ball from the same point onto an inclined ramp, from 10-50 degrees and measured the distance from where the ball was dropped and when it hit the table (i.e. place where it landed after it bounced off the ramp).

    2. Relevant equations

    SUVAT equations?

    3. The attempt at a solution

    So now I need to explain the results, I was thinking that it is due to the component of the velocity, so Vcosθ, changes from 25-40, in such a way that cosθ decreases, therefore the horizontal range of the ball decreases. However I then found that cosθ from 15-40 all decreases, so surely the horizontal range of the ball should decrease. So, I am confused, is my thinking correct or am I missing something? (Maybe V, the speed when the ball hits the ramp is different when the ramp is inclined at different angles?).

    Any help is appreciated! Thanks!
     
    Last edited: Mar 18, 2014
  2. jcsd
  3. Mar 18, 2014 #2
    Did you drop the ball on top of the incline?
    So the ball moved along a vertical line until it touched the plane?
    You need to give more details.
    Was the initial height of the ball always the same? (before letting go)
    And how did you change the angle?

    You may have two effects: the angle of the ball after bumping changes. But the initial height (after collision) may change too. This will result in a change in the magnitude of the initial speed as well.
    Neglecting the effects of this variation in height, the maximum range is obtained for initial speed at 45 degrees in respect to the horizontal. Which means an angle of the plane of 22.5 degrees.
    So your data may be OK.
     
  4. Mar 18, 2014 #3
    Image attached.
     
    Last edited: Mar 18, 2014
  5. Mar 18, 2014 #4
    Interesting problem. I would say you would need to use conservation of momentum and calculate your coefficient of restitution by finding your initial velocity before hitting (easy conservation of energy) and then measuring your rebound velocity, while keeping in mind that your normal force is going to cause the ball to rebound at a right angle to your surface.

    I also think you should be able to calculate your initial rebound velocity by knowing the range of the trajectory and factoring in the extra height drop.
     
  6. Mar 18, 2014 #5
    It only needs to be a simple explanation, so I am wondering why does distance travelled by the ball decrease at a certain θ and above.
     
  7. Mar 18, 2014 #6
    Ok, that makes sense. It seems like the angle at which your maximum trajectory would occur is when your θ = 45°.

    If this seems odd, think about the equation for the range of a trajectory.

    R = V^2sin(2Θ)/g

    What would be angle to maximize this range, considering you aren't changing your velocity nor the acceleration of gravity, and keeping in mind the maximum value of the sine function is 1?
     
  8. Mar 18, 2014 #7
    Ah -- you know what, I am wrong because I missed something. You said that you dropped the ball from the same point? That means that your velocity was increasing as you decreased your angle, because the distance over which it was able to accelerate was increasing. At some angle, your downward velocity and the x component of your velocity was maximized, because once your angle was too shallow, the rebound had too much of a y based component. But, as the theta angle increased, there was not enough distance for your ball to gain a sizeable velocity.

    I hope that helps, and please ask if you need clarification!
     
  9. Mar 18, 2014 #8
    So from your explanation, the angle in which the downward velocity was maximised and therefore the x component is 25. And from this angle onwards, the short distance meant that they downward velocity and x component was less? Is this right?

    Also I am a bit confused by: "because once your angle was too shallow, the rebound had too much of a y based component".
     
  10. Mar 18, 2014 #9
    Yes, you are on the right track!

    The 25° angle is the culmination of two factors being joined together in a maximal way.

    Think about if your angle was 0. What would the trajectory be? 0, right? Because it would merely bounce straight upward, with no horizontal velocity. But, if the angle is 0, your downward velocity will be maximized because if will fall the full height. (Think of the kinematic equations, what is the relationship between V, a, and your distance travelled?)

    Now, if your angle were at the other extreme, you would have a maximal horizontal component, but your ball would never have enough distance over which to accelerate in order to hit a greater velocity.

    If you were to actually give the initial height, you could find the relationship between the impact velocity and the angle of your incline. In fact, this exact relationship could be written as

    V = √(2g(h-BtanΘ)) where h = initial height from ground level, B = the horizontal distance of the ball from the base of your incline, and g is acceleration of gravity.
     
  11. Mar 18, 2014 #10
    Where?
     
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