# Ball bouncing on inclined ramps

• Munchies645
In summary, last week a practical exercise was conducted in which a ball was dropped onto an inclined ramp at various angles (10-50 degrees). The distance between the drop point and the point of impact on the table was measured. The results showed that the horizontal range of the ball decreased as the angle of the ramp increased. Some possible explanations for this phenomenon were discussed, including the component of velocity and the initial height of the ball after collision. The use of conservation of momentum and the coefficient of restitution were suggested as potential methods for further analysis. Additionally, it was noted that the maximum range is achieved when the angle of the ramp is 45 degrees, as supported by the equation for the range of a trajectory. Ultimately, it was concluded that
Munchies645

## Homework Statement

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).

SUVAT equations?

## 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!

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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.

Image attached.

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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.

Rellek said:
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.

It only needs to be a simple explanation, so I am wondering why does distance traveled by the ball decrease at a certain θ and above.

Munchies645 said:
It only needs to be a simple explanation, so I am wondering why does distance traveled by the ball decrease at a certain θ and above.

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?

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.

Rellek said:
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.

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".

Munchies645 said:
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".

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.

Munchies645 said:
Image attached.

Where?

## 1. How does the angle of the ramp affect the height of the ball's bounce?

The angle of the ramp determines the potential energy of the ball. The higher the angle, the higher the potential energy. As the ball rolls down the ramp, this potential energy is converted into kinetic energy, causing the ball to bounce higher.

## 2. Does the material of the ramp affect the ball's bounce?

Yes, the material of the ramp can affect the ball's bounce. A rougher surface may cause more friction, slowing down the ball's momentum and resulting in a lower bounce. A smoother surface may allow the ball to roll more smoothly, resulting in a higher bounce.

## 3. How does the height of the ramp affect the distance the ball travels?

The height of the ramp affects the ball's initial velocity. The higher the ramp, the greater the initial velocity, and therefore the further the ball will travel. However, this assumes a frictionless surface and neglects the effects of air resistance.

## 4. Can the weight of the ball affect its bounce on inclined ramps?

Yes, the weight of the ball can affect its bounce on inclined ramps. A heavier ball will have more potential energy as it rolls down the ramp, resulting in a higher bounce. Additionally, a heavier ball may experience more friction on the surface of the ramp, potentially resulting in a lower bounce.

## 5. How does the shape of the ramp affect the ball's bounce?

The shape of the ramp can affect the ball's bounce in various ways. For example, a steeper ramp will result in a higher potential energy and therefore a higher bounce. A curved ramp may also affect the trajectory of the ball, resulting in a different bounce pattern compared to a straight ramp.

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