Finding velocity upon collision

[Godisnemus]

Homework Statement

A football player (90 kg) moving east collides with another football player (80 kg) moving north. When both players collide, a helmet (2 kg) is ejected south at a velocity of 5 m/s. If both players fall to the ground and slide a distance of 2 m North-East with a friction coefficient of 0.6,

What was the velocity of both football players upon collision?

Homework Equations

[/B]
I'm assuming this requires the Kinetic Energy and the Work Energy Theorem.

W = K
Wf = Kf - Ki
Wf = 0 - Ki

3. The Attempt at a Solution

I have absolutely no idea how to solve this problem. Every single collision example in the class manual or anywhere online for that matter has more data in the problem statement. Either the initial velocity is given for both football players and/or there is orientation angle given to the sliding distance with the friction coefficient.

It would be much appreciated if somebody could point in the right direction to solve this problem. Tutorials or similar examples would also be great. Thank you.

 

[PeroK]

I'm assuming this requires the Kinetic Energy and the Work Energy Theorem.

I'm not sure that's a correct assumption. Can you think of anything else that may be relevant?

 

[TomHart]

North-east means that the angle is 45° north of east.

 

[Godisnemus]

Hi. Thanks for answering.

Not that I can think of. The only formula in the collision section of my book that contains the coefficient of friction (Uc) is that one. :

-Uc (m₁ + m₂) (g) (d) = - [(m₁ + m₂)(v_o²)/2] (d is the sliding displacement)

 

[TomHart]

Have you studied momentum?

 

[PeroK]

Hi. Thanks for answering.

Not that I can think of. The only formula in the collision section of my book that contains the coefficient of friction (Uc) is that one. :

-Uc (m₁ + m₂) (g) (d) = - [(m₁ + m₂)(v_o²)/2] (d is the sliding displacement)

Perhaps this question requires you to take two different things that you've learned and apply them to different parts of the problem?

1) You have a collision.

2) You have sliding with friction.

If you tackle the second part using what you know about friction, what does this give you?

 

[Godisnemus]

Not at this point. But it wouldn't surprise me if I needed to use concepts that I haven't specifically learned yet because it is a bonus question.

So just to make things clear, I should not be using the equation below in order to determine the velocity of the collision:

Wf = 0 - Ki

Uc (m_player1 + _mplayer2 - m_helmet) (g) (d) = - [(m₁ + m₂)(v_o²)/2]

 

[Godisnemus]

Given (V₀) from the equation above, I would then use:

Σp = m₁v₁ + m₂v₂= (m₁ + m₂) v₀

Σp_x = m₁v₁ + 0 = (m₁ + m₂) v₀ cos 45° (Velocity of player 1)

Σp_y = 0 + m₂v₂ = (m₁ + m₂) v₀ sin 45° (Velocity of player 2)

 

[PeroK]

Not at this point. But it wouldn't surprise me if I needed to use concepts that I haven't specifically learned yet because it is a bonus question.

So just to make things clear, I should not be using the equation below in order to determine the velocity of the collision:

Wf = 0 - Ki

Uc (m_player1 + _mplayer2 - m_helmet) (g) (d) = - [(m₁ + m₂)(v_o²)/2]

For the second part, you can use either conservation of energy or a kinematic equation. But, in either case the ejected helmet should be subtracted from both masses.

Given (V₀) from the equation above, I would then use:

Σp = m₁v₁ + m₂v₂= (m₁ + m₂) v₀

Σp_x = m₁v₁ + 0 = (m₁ + m₂) v₀ cos 45° (Velocity of player 1)

Σp_y = 0 + m₂v₂ = (m₁ + m₂) v₀ sin 45° (Velocity of player 2)

What about the helmet?

IMO, this is a poor question. Does the 80kg and 90kg include the mass of a helmet each? What is the combined mass of the two players after the collision? It could be 172kg, 170kg or 168kg.

My guess is 168kg.

 

[Godisnemus]

I assumed that when both players collided and the helmet went flying off, the total mass would be 80kg + 90kg - 2kg as they were sliding on the ground. I also assumed that both 80kg and 90kg are the masses of each player fully equipped.

You're right. I forgot to take into consideration the helmet for that part. Like I said, I find this question to be very "outside the box". I've tried very hard to solve it and I am just about to give up on it.

 

[PeroK]

I assumed that when both players collided and the helmet went flying off, the total mass would be 80kg + 90kg - 2kg as they were sliding on the ground. I also assumed that both 80kg and 90kg are the masses of each player fully equipped.

You're right. I forgot to take into consideration the helmet for that part. Like I said, I find this question to be very "outside the box". I've tried very hard to solve it and I am just about to give up on it.

I think your assumption is correct. It should say that though.

I think you are nearly there. The helmet is just an adjustment to the equation for y-momentum

 

[Godisnemus]

Thanks for your input man, much appreciated.

I'll bring my attempt at the solution to a TA and see what they have to say about it.

 

[TomHart]

Given (V0) from the equation above, I would then use:

vo from what equation? I believe that you can use the first equation you listed (shown below).

-Uc (m1 + m2) (g) (d) = - [(m1 + m2)(vo2)/2]

That appears to be equating [the work done to stop the two sliding objects] to [the initial kinetic energy of those two objects] - initial kinetic energy meaning after they collided, not before they collided. But you can't include the helmet as part of that because it did not slide with the two players.

I don't know if you have studied elastic and inelastic collisions, but this problem is an inelastic collision because the 2 objects (football players) stuck together after impact. (A perfectly elastic collision would be similar to two billiard balls colliding. In a perfectly elastic collision, the total kinetic energy is the same before and after the collision.) So, in your problem, the total kinetic energy immediately before the collision is not equal to the total kinetic energy immediately after the collision.

Once you know the initial velocity of the two players (from your equation), you should be able to work the solution using x and y components of momentum.