# Does the make sense (momentum conservation)

1. Apr 9, 2009

### ashvuck101

1. The problem statement, all variables and given/known data

A hockey puck moving at a speed V1 collides head on with a second identical puck moving toward it at speed V2. After the collision, the first puck slows down to speed v1 without changing direction. (5 marks)
a. After the collision what is the speed v2 of the second puck?
b. Calculate the speed v2 of the second puck when the first puck had an initial speed of 18ms-1 that was changed to 2.0ms-1 by the collision and the initial speed of the second puck was 12.0 ms-1. Both pucks have a mass of 0.16kg.
c. Does your answer change if the masses of both pucks are doubled?

2. Relevant equations

my tutor gave me this equation

m(v1-v2)=m(u1-u2)

but i dont know how he got it
3. The attempt at a solution

m(v1-v2)=m(u1-u2)

U2= v1-v2+u2
=2-18+12
=-4

but that be right cuz the puck does not change direction???

Last edited: Apr 9, 2009
2. Apr 9, 2009

### Chewy0087

Okay, well you know! that momentum is conserved right? Now let's think about what that might mean in an equation.

Momentum = mass * velocity right?

So if two things are going in opposite directions

M1 * V1 - M2 * U1 = the momentum of the overall state (m1 bieng mass of the 1st, v1 bieng velocity of the first, and the similiar routine with m2 bieng mass of the second and u1 bieng velocity of the second)

Because momentum is conserved, this must be equal to;

M1 * V2 - M2 * U2 (with v2 & u2 bieng the new speeds after the collision)

So okay, let's put them two together;

M1 * V1 - M2 * U1 = M1 * V2 - M2 * U2

Now let's get M1's and M2's on the same side;

M1 * V1 - M1 * V2 = M2 * U1 - M2 * U2

Now take the m1's and m2's out as factors. - That's how that equation was reached, i hope you understand it, it's key, but personally i don't like remembering that equation and just thinking about it inmy head. As far as your question goes;

a) Think about it, it must be v2, even though it's a horribley worded question...where did you get it from?! :P

b) You're right

c) you're right - no matter how heavy the m's they just cancel

You're doing the working bang on :P

3. Apr 9, 2009

### Staff: Mentor

Too many minus signs to decipher. Just use plain old conservation of momentum:

mv1 + mv2 = mv'1 + mv'2

Be careful with signs when you substitute the given speeds into the equation. (For example, v2 = -V2.)

4. Apr 9, 2009

### Chewy0087

I think that can be confusing Al, but yeah you're right.

Its worth stressing though that because in your example it's going in the opposite direction the velocity must be negative, but in general it is +'s rather than minus', sorry if my example was misleading.

5. Apr 9, 2009

### ashvuck101

the answer still cant be right becuase the puck cant change direction though that is what confuses me so i did it my deviding the ratio of he intial and final velocity (too hard to write down) and solved for x and got 3 ms

6. Apr 9, 2009

### Chewy0087

? :P

Of course the second puk changes direction - think about it logically, 2 things going at each other head on, ONE of them has to change direction, since the first one hasn't, the second one must....

Edit : Calculation (sorry if im diong this for you)

Momentum before;

M*18 - M12 = M2 + MX It's a plus because it HAS changed direction, now / by X

6 = 2 + X

Yep, you were right, 4.

7. Apr 9, 2009

### ashvuck101

yeah i guess you a right i think i was confused by the wording of the question and the fact that speed cant be a negative value

i guess i can say the speed is 4 m/s in the opposite direction since the whole system moves in the direction of the first puck

8. Apr 9, 2009

### ashvuck101

wait i was right what did i do wrong in my calculations???

how did you get a postive value instead of a negative value?

9. Apr 10, 2009

### Chewy0087

m(v1-v2)=m(u1-u2)

cancel the mass's

v1 - v2 = u1 - u2
18 - 2 = 12 - -X

there must be another - before the X because it has changed direction, so you get

16 = 12 + X

X = 4 m/s.

However that forumula can be confusing and again i stress it's better to think about it than using formula's such as these, which are specific to problems and hard to get your head around!