# Completely lost

1. Mar 4, 2004

### Mac13

completely lost:(.. plz help

Ok I have been looking at this problem and i really don't know where to start. I've done all the other homework in my section but cannot get this. I would appreciate it if someone would help me:). Here it is.

Two particles, of masses m and 2.88m are moving toward each other along the x axis with the same initial speeds 5.81m/s. Mass m is traveling to the left and mass 2.88m is traveling to the right. They undergo an elastic glancing collision such that m is moving downard after the collision at a right angle to its initial direction.

1.Find the final speed of the 2.88m mass.

2.What is the angle at which am is scattered. Answer between -180 and +180.

2. Mar 5, 2004

Since this collision is elastic, you know two things: momentum is conserved, and energy is conserved.

So write out the equations for momentum before and after the collision, and equate them. Then write out the equations for energy before and after the collision and equate those. Solve the system of equations.

3. Mar 5, 2004

### Mac13

i dotn know what formulas to use...when u said that i went and looked under the review for the chapter and found a formula for an elastic collision

final v2= ((2m1*V1i)/m1+m2) + ((m2-m1)*v2i)/m1+m2)

where m1 is mass 1...m2 is mass 2....i = initial

..i calculated this and the dumb website said i was wrong:\

4. Mar 5, 2004

Well, if you don't understand a formula, then it doesn't help to use it, does it? Stop trying to skip steps. It's just making your life harder.

If you don't know the formulas for momentum and for kinetic energy, then you're going to really be hating life for the rest of the semester. I suggest remembering them; it's really not too hard.

Momentum:
p = m*v

Kinetic Energy:
$$KE = \frac{1}{2}mv^2$$

Now start from the beginning--that is, conservation of momentum and conservation of energy--and use these formulas to get a system of equations. Then solve that system. I'll even get you started.

Conservation of momentum:
$$p_{x,1,b} + p_{x,2,b} = p_{x,1,a} + p_{x,2,a}$$
$$m_1v_{x,1,b} + m_2v_{x,2,b} = m_1v_{x,1,a} + m_2v_{x,2,a}$$
$$(m) (-5.81\frac{\textrm{m}}{\textrm{s}}) + (2.88m)(5.81\frac{\textrm{m}}{\textrm{s}}) = (m)(0) + (2.88m)v_{2,a}$$

Now repeat this for the y components of momentum and kinetic energy.

5. Mar 5, 2004

### Mac13

ya i remembered those formulas just didnt know what to do with those..thought it was something more complex..and whats the a and b subscripts?

6. Mar 5, 2004

### Mac13

after using the quadratic equation i got V2 = 6.6435 and -0.1803 but the 6.6 one didnt work and i didnt think it was the negative one

7. Mar 5, 2004

The a and b subscripts denote after collision and before collision, respectively.

Go with the answer that makes physical sense.

8. Mar 5, 2004

### Mac13

i went with the 6.6 one and the dang site said its wrong:/...ohhh..would it be negative then?..hehe

9. Mar 5, 2004

Show me what you've done and we can see where you went wrong.

10. Mar 5, 2004

### Mac13

alright..

1/2mv1^2 + 1/2(2.88m)v2^2 = 65.4867

v1 = 10.9228 - 2.88v2

plugging v1 into that first equation i got
1/2m(10.9228-2.88v2)^2 + 1.44v2^2 = 65.4867
1/2m(8.2944v2^2-62.9153+119.3075)+1.44v2^2 = 65.4867
right here i multiplied both sides by 2 to get the 1/2 out but i didnt know what to do with the m so i just acted like it was 1(??? lol)

then finishing all that up with the quad. formula i got v2 = 6.6435 or -0.1803

11. Mar 5, 2004

Where'd the second equation you used come from?

12. Mar 5, 2004

### Mac13

13. Mar 5, 2004

### Mac13

i followed that..its down at bottom