Change in Momentum of a Biker: Homework Solution

[temaire]

Homework Statement

A biker, who is 72.4 kg, rides down a hill at 4.76 m/s $$[3.83^\circ$$ down from south], when he hits a bump which makes him go 2.00 m/s $$[1.80^\circ$$ up from south]. What is the change in momentum?

Homework Equations

$$p=mv$$

The Attempt at a Solution

Here's what I did:

p=(72.4)(2.00+4.76)
p=489 kg m/s

I know this is probably wrong, but I didn't know how to accommodate the degrees. Can someone help me?

 

[Doc Al]

Momentum and velocity are vectors, so direction counts. I suggest that you find the horizontal and vertical components of the velocity before and after he hits the bump, then use those to find the change in velocity.

 

[temaire]

So is this what I have to do?

And then the answer would be 490 kg m/s [$$2.17^\circ$$ down from south]?

 

[Doc Al]

Your first two diagrams, where you found the components of the velocities, are perfect. But to find the change in velocity you need to subtract the two vectors, which means subtracting the components. Change = Final velocity - initial velocity. So do that part over and you should be OK. (Pay attention to the signs of the components; let downward = negative.)

 

[temaire]

So you're saying that I must subtract the final sum of the components? But how is it possible to subtract x and y components when they are considered separately? I've never done this before in class.

 

[Doc Al]

To find the components of the change in velocity, you subtract the components of the final and initial velocities. You subtract the x and y components separately, not their sum.

 

[temaire]

Oh ok, I understand what you're trying to say. Just give me a minute, I'll fix my work, and post it.

 

[temaire]

Here's the new work:

So is the answer 201 kg m/s [$$7.88^\circ$$ up from north]?

 

[Doc Al]

Perfect!

 

[temaire]

Thanks Doc Al, you've been a great help.