# Find the velocity of each object after the collision

1. Jun 28, 2010

### mandy9008

1. The problem statement, all variables and given/known data
A 28.0 g object moving to the right at 19.0 cm/s overtakes and collides elastically with a 5.0 g object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision.

2. Relevant equations
mv + mv = mv + mv

3. The attempt at a solution
(28.0 g)(19.0 cm/s) + (5.0 g)(15.0 cm/s) = (28.0 g)v1f + (5.0 g)v2f
607 g cm/s = 28 g v1f + 5 g v2f

607 cm/s = 28 v1f + 5 v2f
(574 cm/s = v1f + v2f )-5

607 cm/s = 28 v1f + 5 v2f
-2870 cm/s = 5 v1f - 5 v2f

-2263 cm/s = 23 v1f
v1f = -98.4 cm/s

v2f = -5.83 cm/s

I know that both velocities have to be positive, so I know these values cannot be right.

2. Jun 28, 2010

### rock.freak667

Good, though you should convert your units.

Where did you get these numbers from?

3. Jun 28, 2010

### mandy9008

the reason that I did not convert my units was because the answer asks for velocities in cm/s.

607 g cm/s = 28 g v1f + 5 g v2f I solved the mv + mv = mv + mv equation for with the 2 unknowns.

then I subtracted 33 from each side to set v1f and v2f equal to 1, and i got 574 cm/s = v1f + v2f

then I multiplied the bottom equation by 5 so I could cancel out the vf2 so I could solve for vf1, thus getting -2870 cm/s = 5 v1f - 5 v2f

after that, I added the 2 equations together and got -2263 cm/s = 23 v1f, then i solved for v1f: v1f = -98.4 cm/s

then I took the v1f that I got and plugged it into the original equation and got v2f: v2f = -5.83 cm/s

4. Jun 28, 2010

### rock.freak667

I am still perplexed as to how you solved one equation with two unknowns.

5. Jun 28, 2010

### mandy9008

okay, how would you have done this problem. It is hard to explain how I did it without talking in person

6. Jun 28, 2010

### rock.freak667

Collision: conservation of momentum
Elastic: Conservation of energy

Two equations, two unknowns.

7. Jun 28, 2010

### mandy9008

but i have two unknowns. I am confused

8. Jun 28, 2010

### rock.freak667

If you are talking about what you did, then I too am confused.

If you meant what I suggested; then each conservation law would give an equation

You end up with something like

ax+by=e1
cx+dy=e2

Which you can solve for both x and y.

9. Jun 28, 2010

### inky

Hello Mandy,

You said subtracted 33 from each side but v1f and v2f are not the same you cannot do like this.

You should use law of conservation of momentum and law of conservation of kinetic energy to solve 2 unknown problem.

10. Jun 28, 2010

### mandy9008

what are your a's and b's and such?

11. Jun 28, 2010

### rock.freak667

a,b,c,d,e1,e2 are just random constants, I am just showing you how you'd get two equations to solve. (One will not be linear though)

12. Jun 28, 2010

### mandy9008

I am way more confused then when I started. My prof did a problem similar to this one in class today and he did it the way that I did it, but my answer is not coming out right. UGH

13. Jun 28, 2010

### inky

Here your first equation, it is correct.
607 g cm/s = 28 g v1f + 5 g v2f

Second equation is
Before collision total KE= After collision total KE

14. Jun 28, 2010

### rock.freak667

Then let's start again, but without doing anything major.

The problem says the masses collide elastically.

Collision means that you will need to apply conservation of linear momentum.
Elastic means you will need to apply conservation of kinetic energy.

This is what you get when you applied conservation of momentum.

Kinetic energy of a mass is given by 1/2mv2.

So kinetic energy before = kinetic energy after. So form this equation now.

15. Jun 28, 2010

### mandy9008

1/2 mv21i + 1/2 mv22i = 1/2 mv21f + 1/2 mv22f

1/2 (28g)(19 cm/s)2+ 1/2 (5 g)(15 cm/s)2 = 1/2 (28g)v21f + 1/2 (5g)v22f

16. Jun 28, 2010

### inky

Yes, right.
You have previous for momentum. You can easy to get the answer.