Inelastic Collision Problem

1. Nov 25, 2013

ryanx806

1. The problem statement, all variables and given/known data
A halfback with a mass m1=90kg is running up field with a speed of v1=7.6m/s. He is tackled by an opponent with a mass of m2=120kg, who approaches the halfback at an angle of 30 degrees with a velocity of v2=4.0m/s. Assume the collision is perfectly inelastic.

a) Calculate the magnitude and direction of the resultant velocity of the two players just after the tackle.

b) Calculate the change in kinetic energy (in Joules) for the system of the two players.

2. Relevant equations
I'm confused about calculating the initial momentum of each of the players, and how to add them together to get the final velocity and direction.

3. The attempt at a solution

2. Nov 25, 2013

vela

Staff Emeritus
Remember that momentum is a vector, so it has a magnitude and direction. Draw a sketch of the momenta before the collision. Let's orient the axes so that the halfback is running in the +x direction. You can represent his momentum as an arrow pointing to the right. How long should this arrow be? Which way should the arrow representing the opponent's momentum be? What is its magnitude?

To combine them, you have to add them vectorially. That means breaking them up into components and adding the respective components together.

3. Nov 25, 2013

ryanx806

My professor gave us an example that uses the equation for conservation of momentum (mvi=mvf), which I understand. But I'm confused on whether I need to solve this equation for both players in both axes (which would give me 2 final x-velocities and 2 final y-velocities) or if I just use the equation once to find a total final x- and y-velocity.

4. Nov 25, 2013

mic*

Does the "perfectly inelastic collision" part tell us anything important about the outcome?

5. Nov 25, 2013

ryanx806

So...

(90kg)(7.6m/s) + (120kg)(4m/s) = 1164kg*m/s

Does that mean that the final velocity = 1164/(90kg + 120kg) ?

And if so, how do I know which direction they are going after the collision?

6. Nov 25, 2013

mic*

I am sorry. I deleted the equation line I wrote because I realised it may be a little misleading.

You need to add them as vectors like vela explained. Letting the halfback's momentum be in the x(+ve) direction is a sensible place to start. Drawing out the vectors may be helpful too.

Eg. Halfback's momentum => (90)*(7.6) = 684
Therefore coordinates of this vector (starting from the origin) are:
x = 684
y = 0

Apply trigonometry (using the angle given in the problem) to the magnitude of the second vector to get its x and y components. Be sure to remember if the components have negative values. Eg if the opponent were running straight at the halfback, his x component would be negative because we set the halfbacks motion to be in the positive x direction.

Once you have x and y components for each players momentum vector, add the two x components, and the two y components. This will give you the x and y components of the final vector. From there you can use pythagorus and trig to get the magnitude and direction of the final congealed mass of tangled players, respectively.

7. Nov 25, 2013

ryanx806

Okay! Thank you so much! I think that's what I was originally trying to do, but my professor's example made me question it. I really appreciate your help though!