Momentum of a ball bouncing off of the floor

[FireKyuubi]

Homework Statement

A stress ball that has a mass of 300 g . The ball is thrown towards the floor at an angle. Right before hitting the floor, the ball has a speed of 11.0 m/s and its trajectory makes an angle 55 º above the floor. However, after the collision, the ball bounces back from the floor with an angle 25 º from the floor. If the ball was in contact with the floor for 250 ms , what was the average force exerted on the ball by the floor?

Relevant Equations

$m\vec v_1 - m\vec v_2 = \vec Ft$

$m=.3 kg, v = 11 \frac {m}{s}, t = .25 s, \vec v_1 = mv\langle cos(-55), sin(-55) \rangle, \vec v_2 = mv\langle cos(25), sin(25) \rangle$
$$m\vec v_1 - m\vec v_2 = \vec Ft = mv\langle cos(25)-cos(-55), sin(25)-sin(-55) \rangle$$
$$Ft = mv\sqrt{(cos(25)-cos(-55))^2 + (sin(25)-sin(-55))^2}$$
$$F = 16.97 N$$
The answer is supposed to be 14.3 N, but I don't understand what's wrong with the equation I've set up.

 

[PeroK]

Can you start by explaining all the assumptions you have made? What happens when the ball hits the floor?

Note that the question itself ought to say something about these assumptions. I guessed what the question setter intended, but that was an educated guess.

Note that the question is fundamentally flawed in another way as it assumes a constant force. The force itself would be harmonic and the average would be much harder to calculate - a lot of question setters make this casual mistake of using "average force" instead of "assume constant force".

 

[FireKyuubi]

Can you start by explaining all the assumptions you have made? What happens when the ball hits the floor?

Note that the question itself ought to say something about these assumptions. I guessed what the question setter intended, but that was an educated guess.

Note that the question is fundamentally flawed in another way as it assumes a constant force. The force itself would be harmonic and the average would be much harder to calculate - a lot of question setters make this casual mistake of using "average force" instead of "assume constant force".

Oh, sure, I'm assuming that when the ball hits the floor, there will be some impulse that causes the ball's momentum to change. And that impulse divided by time is average force, which should be a constant vector. Taking the magnitude of that constant vector will give me the answer.

 

[PeroK]

Oh, sure, I'm assuming that when the ball hits the floor, there will be some impulse that causes the ball's momentum to change. And that impulse divided by time is average force, which should be a constant vector. Taking the magnitude of that constant vector will give me the answer.

What have you assumed happens to the speed of the ball before and after the collision?

In particular, what have you assumed about the horizontal and vertical components of the velocity before and after?

Hint: compare your horizontal component before and after it bounces.

 

[Orodruin]

Note that the question is fundamentally flawed in another way as it assumes a constant force. The force itself would be harmonic and the average would be much harder to calculate - a lot of question setters make this casual mistake of using "average force" instead of "assume constant force".

I disagree. Since impulse is the time integral of the force, the time average is exactly impulse divided by time - which is the same as you obtain by assuming a constant force.

 

[PeroK]

I disagree. Since impulse is the time integral of the force, the time average is exactly impulse divided by time - which is the same as you obtain by assuming a constant force.

Of course, the average force is easy to calculate!

 

[FireKyuubi]

What have you assumed happens to the speed of the ball before and after the collision?

In particular, what have you assumed about the horizontal and vertical components of the velocity before and after?

Hint: compare your horizontal component before and after it bounces.

I assumed that the speed would stay constant, but since there's impulse that doesn't have to be the case. There's more horizontal velocity after the bounce than before, is that what you mean?

 

[haruspex]

I assumed that the speed would stay constant, but since there's impulse that doesn't have to be the case. There's more horizontal velocity after the bounce than before, is that what you mean?

Ignoring any friction or spin, what direction is the impulse in?
Don't assume mechanical energy is conserved.

 

[FireKyuubi]

Ignoring any friction or spin, what direction is the impulse in?
Don't assume mechanical energy is conserved.

Its pointing to the right and up, decreasing vertical velocity and increasing horizontal velocity.

 

[haruspex]

Its pointing to the right and up, decreasing vertical velocity and increasing horizontal velocity.

Why would it point that way?
When two objects come into contact without friction, the force between them is the minimum necessary to prevent interpenetration.

As I wrote, don’t assume mechanical energy is conserved, so don’t assume speed is conserved.

 

[FireKyuubi]

Why would it point that way?
When two objects come into contact without friction, the force between them is the minimum necessary to prevent interpenetration.

As I wrote, don’t assume mechanical energy is conserved, so don’t assume speed is conserved.

So it's pointed directly up?

 

[PeroK]

So it's pointed directly up?

That's what I assumed and got the given answer.

 

[FireKyuubi]

That's what I assumed and got the given answer.

So is impulse always perpendicular to the surface, like the normal force?

 

[haruspex]

So is impulse always perpendicular to the surface, like the normal force?

Yes, unless there is friction and the object has moment of inertia.