Speed and direction of Velocity Vector - Inelastic

[garretts16]

Homework Statement

A 55kg man is running in the positive y direction at 4 m/s. He is tackled by a 115kg man running 8.5 m/s at a 140 degree angle. What is the speed and direction after the inelastic collision?

Homework Equations

p=mv
p = sqrt of x^2 + y^2

The Attempt at a Solution

The final velocity is 7.04 m/s, but the angle keeps coming out at 130 degrees, which I know is wrong.

 

[TSny]

Hello.

p = sqrt of x^2 + y^2

What do the x and y stand for here?

The final velocity is 7.04 m/s, but the angle keeps coming out at 130 degrees, which I know is wrong.

Can you please show steps for how you got these answers? Then maybe we can track down where you're having difficulty. It will also help if you state what physics principle(s) you are using.

When you say that the angle is 140 degrees, that's not really very clear. Is the 140 degrees measured counterclockwise from the +x axis?

 

[garretts16]

the x is the velocity in the x direction, and same for y

I multiplied each mass times the velocity to get momentum. Then I set that P = MV1 + MV2 and I solved for the missing V. The angle is 140 degrees from 0. Counterclockwise.

 

[TSny]

the x is the velocity in the x direction, and same for y[\QUOTE]
Should the mass also be in there somewhere?
[\QUOTE]
I multiplied each mass times the velocity to get momentum. Then I set that P = MV1 + MV2 and I solved for the missing V. The angle is 140 degrees from 0. Counterclockwise.

You still haven't shown much detail. How did you use the angle of 140 degrees? Did you only set up one equation to solve for V?

 

[garretts16]

I haven't used the angle yet, I'm not sure how to use it to find the direction after the collision. I set up a momentum equation, and, since it's an elastic collision, I set up a KE equation with it. I had trouble when I ran through the equations: I solved for x in the MV equation, but I got a y value in the answer. Now I'm supposed to plug it into the KE equation for it's x, then solve to find two possible answers.

 

[TSny]

The collision is given to be inelastic, not elastic. So, you won't be able to use an energy equation to help you find the final velocity.

Momentum is a vector quantity. It has both x and y components: p_x and p_y. When momentum is conserved, the total x component of momentum will be conserved and the total y component of momentum will be conserved. So, you can set up two equations: one equation for conservation of the x component of momentum and a separate equation for the y component of momentum.

So, to get started, see if you can find the following:

1. the x-component of momentum of the 55 kg person before the collision

2. the x-component of momentum of the 115 kg person before the collision

3. the y-component of momentum of the 55 kg person before the collision

4. the y-component of momentum of the 115 kg person before the collision

You're going to need to use that 140 degrees to get answers for 2 and 4.

 

[garretts16]

1. 0
2. -6.5
3. 4
4. 5.46

from what I gathered.

 

[garretts16]

Actually, I found the components of the velocity by mistake.

 

[garretts16]

1. 0
2. -748.8
3. 220
4. 628.32

 

[TSny]

OK. Good. Suppose you let V_x be the x-component of the final velocity and V_y the y-component of the final velocity.

What would the equations for conservation of the x and y components of momentum look like?

 

[garretts16]

I don't fully understand the question.

 

[garretts16]

I think I solved it on my own. Is the new angle 131.4 degrees from the +x ccw?

 

[TSny]

I think I solved it on my own.

Great!

Is the new angle 131.4 degrees from the +x ccw?

Yes, I think so.