Ball hitting a wall at an angle - x and y motion

In summary: If the wall was frictionless, the y-motion graphs would look no different than if the ball did not hit a wall, correct? The y vs t graph would resemble a parabolic curve, v vs. t: a line with a negative slope intersecting zero at a point corresponding with the apex of the parabola, and the acceleration would be negative (-9.8). The graph should look the same no matter what. However, if the wall is frictionless then the ball will not bounce back.
  • #1
peesha
6
0
Hello! Hopefully this will be a fairly straightforward conceptual problem.
A heads up - while this problem clearly deals with momentum and elastic collisions, we haven't actually covered these concepts in class yet. I'm getting the idea, but am still a bit hazy here and there.


Homework Statement


A ball bounces off a wall. The ball is traveling at an angle 10 degrees below the horizontal when it hits the wall. The friction force by the wall is equal to the weight of the ball.

For the sake of clarity, let's say that the ball was thrown from left to right.

We are asked to:
1)Draw a free body diagram of the ball during contact with the wall
2)Sketch all 6 x-motion and y-motion graphs - that is, x/y vs t, velocity vs t, and acceleration vs t. The graphs must specifically identify three time periods: before, during, and after contact.


2. The attempt at a solution

The free body diagram seems pretty simple. The vertical forces are the force due to gravity (down) and friction force (up). These forces, as stated in the problem, are opposite and equal. There is one horizontal force: the normal force of the wall on the ball. The force is perpendicular to the wall (The normal force would be perpendicular regardless of the angle of impact, correct?) There is a net force in the leftward direction and no net vertical force while the ball is in contact with the wall.

As for the motion diagrams: I have no problem with the x-motion. The x vs t graph will increase steadily, be horizontal while the ball is in contact with the wall, then decrease steadily at the same rate. The v vs t graph will be horizontal and positive before contact, horizontal and negative after, and have a negative (but not linear) slope during contact. The a vs t graph will be zero except for a negative 'blip' during contact.

The y-motion graphs I'm having a bit more trouble with. If the wall was frictionless, the y-motion graphs would look no different than if the ball did not hit a wall, correct? The y vs t graph would resemble a parabolic curve, v vs. t: a line with a negative slope intersecting zero at a point corresponding with the apex of the parabola, and the acceleration would be negative (-9.8).

My big question is what happens when the ball is in contact with the wall if the net vertical force is zero during contact (due to friction)? It seems like there will be a very short 'break' in the parabola of the y vs. t graph, where ∆y is zero. What about the v vs. t graph? In this case, it seems like there would be two points in the graph that intersected zero. A vs t would also have to reach zero for a moment.

Any guidance on the y-motion would be much appreciated.

Thanks so much for any help.
 
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  • #2
Is the ball to be taken as a point mass? No radius is mentioned, so I assume so. (If not, it will get quite complicated as the ball will acquire spin while in contact with the wall.)
peesha said:
The friction force by the wall is equal to the weight of the ball.
That statement is interesting. It would follow that there is no vertical acceleration while the ball is in contact with the wall. If the contact time is taken to be non-infinitesimal that would imply it slides down the wall while in contact. OTOH, if the contact time is infinitesimal then gravity becomes irrelevant. The normal 'force' has to be treated as an impulse - a very large force acting for a very short time to provide a medium momentum change. If the coefficient of friction is nonzero, it follows that the frictional force can also be large. In principle, the ball would bounce back along the line of impact (initially).
So we're thrown back to the sliding case.
There is one horizontal force: the normal force of the wall on the ball. The force is perpendicular to the wall (The normal force would be perpendicular regardless of the angle of impact, correct?)
Yes.
As for the motion diagrams: I have no problem with the x-motion. The x vs t graph will increase steadily, be horizontal while the ball is in contact with the wall,
If contact takes nonzero time then I would imagine the ball is compressed then decompressed. This is basically simple harmonic motion, so I would expect a sine wave here.
then decrease steadily at the same rate.
Unless it is perfectly elastic, it will be slower.
The y-motion graphs I'm having a bit more trouble with. If the wall was frictionless, the y-motion graphs would look no different than if the ball did not hit a wall, correct? The y vs t graph would resemble a parabolic curve, v vs. t: a line with a negative slope intersecting zero at a point corresponding with the apex of the parabola, and the acceleration would be negative (-9.8).
Yes
My big question is what happens when the ball is in contact with the wall if the net vertical force is zero during contact (due to friction)? It seems like there will be a very short 'break' in the parabola of the y vs. t graph, where ∆y is zero.
∆y won't be zero. What will be?
 
  • #3
Thanks so much for you reply!

haruspex said:
Is the ball to be taken as a point mass? No radius is mentioned, so I assume so. (If not, it will get quite complicated as the ball will acquire spin while in contact with the wall.)

Yep. Point mass.

That statement is interesting. It would follow that there is no vertical acceleration while the ball is in contact with the wall. If the contact time is taken to be non-infinitesimal that would imply it slides down the wall while in contact. OTOH, if the contact time is infinitesimal then gravity becomes irrelevant. The normal 'force' has to be treated as an impulse - a very large force acting for a very short time to provide a medium momentum change. If the coefficient of friction is nonzero, it follows that the frictional force can also be large. In principle, the ball would bounce back along the line of impact (initially).
So we're thrown back to the sliding case.

Huh. Because there is vertical velocity. I forgot there was still vertical velocity (I made the jump from a=0 to v=0). But if the ball is hitting the wall at an angle below the horizontal (not yet at it's apex), it would...slide up the wall? (Since velocity is still positive at the point?)

I'm presuming time is treated as non-infinitesimal - simply because we're asked to address the motion/velocity/acceleration of the ball while in contact with the wall.

Unless it is perfectly elastic, it will be slower.

It's acceptable to assume that the bounce is perfectly elastic.

Yes ∆y won't be zero. What will be?

Ok, so ∆y will be a straight line with a positive slope during contact. Velocity will be a horizontal positive line during contact, and acceleration will be zero. How does that sound?

Thanks again. Very helpful!
 
  • #4
peesha said:
if the ball is hitting the wall at an angle below the horizontal (not yet at it's apex),
It says: "The ball is traveling at an angle 10 degrees below the horizontal".
If its path is below the horizontal (which is how I read it) it's angled down.
You're reading it as below the horizontal from the wall's perspective. That may be what was intended, but it's not what it says.
Ok, so ∆y will be a straight line with a positive slope during contact. Velocity will be a horizontal positive line during contact, and acceleration will be zero. How does that sound?
Apart from the uncertainty of whether it's going up or down, fine.
 
  • #5
Ah. Great. I'll clear up that distinction with my professor. Thanks again!
 
  • #6
What will be the momentum of body after colliding elastically with wall?
 
  • #7
Maha Aftab said:
What will be the momentum of body after colliding elastically with wall?
If it is perfectly elastic, what can you say about the horizontal velocities before and after?
Since the frictional frictional force equals the weight of the ball, what can you say about the change in vertical velocity?
 

Related to Ball hitting a wall at an angle - x and y motion

1. What causes a ball to bounce off a wall at an angle?

When a ball hits a wall, it experiences a change in momentum and direction due to the force of impact. This causes the ball to bounce off the wall at an angle that is determined by the angle at which it hit the wall and the surface of the wall itself.

2. How does the angle at which the ball hits the wall affect its motion?

The angle at which the ball hits the wall affects both the horizontal and vertical components of its motion. The horizontal component remains the same, while the vertical component changes direction and magnitude based on the angle of impact.

3. What factors can influence the trajectory of a ball bouncing off a wall?

The trajectory of a ball bouncing off a wall is influenced by several factors, including the initial angle of impact, the elasticity of the ball and wall, air resistance, and any external forces acting on the ball.

4. Why does a ball lose energy when it bounces off a wall?

When a ball collides with a wall, some of its kinetic energy is transferred to the wall as it deforms and then returns to its original shape. This results in the ball losing energy and bouncing back with a slightly lower velocity than before.

5. How can mathematical equations be used to calculate the motion of a ball bouncing off a wall at an angle?

Mathematical equations, such as the laws of motion and conservation of energy, can be used to calculate the motion of a ball bouncing off a wall at an angle. By considering the initial conditions and the forces acting on the ball, these equations can accurately predict the trajectory and velocity of the ball after the collision.

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