# Momentum problem

1. Dec 14, 2013

### Lacnez

A manufacturer is designing a two part firecracker with a total mass of 1.0 kg. The firecracker will launch at a speed of 25 m/s, then separate into its parts at some time, t, after launch. The parts have equal mass. If the parts separate such that the angle of deviation off the original flight path is 7.5° and the speed of the parts is still 25 m/s in the same directions, calculate the change in momentum of the parts. Assume air friction is negligible.

I tried out the question but I don't get far cause I really have no clue on how to do this question.

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2. Dec 14, 2013

### haruspex

Not sure I understand the question. I think it means that each part has the same component of velocity in the original direction before and after separation. If so, what is the direction of the change in momentum of one part? Remember that momentum is a vector.

3. Dec 14, 2013

### Lacnez

I attached the pic that came with the problem maybe that will help the understanding because I cannot make heads or tails with this problem

4. Dec 14, 2013

### Staff: Mentor

Please show your work then, so we can see where you got stuck.

5. Dec 14, 2013

### Lacnez

Ok first I calculated the momentum of the original firecracker

p=mv
p=(1)(25)
p=25 kg m/s

then I took the one of the pieces of the firecracker and split it into its components

Vertical= 25cos7.5
= 24.786 m/s
Horizontal= 25sin7.5
= 3.263 m/s

From here i didn't know what to do so I calculated the momentum of one of the pieces

p=mv
p=(0.5)(25)
p= 12.5 kg m/s

This is as far as I got, I didn't know what to do next.

6. Dec 14, 2013

### Staff: Mentor

The original flight direction is directly upwards (actually it does not matter, but it is easier to calculate if you use the original flight direction as an axis). There is nothing with 7.5° before the split happens.

You don't know the velocity of the pieces here.

Start with the vertical direction. What is the original momentum in vertical direction? What has to be the total momentum in this direction afterwards? There is a conservation law...

7. Dec 14, 2013

### Lacnez

the original vertical momentum is the 25 kg m/s and I'm assuming because of the conservation law that the vertical component of the one of the two pieces will also have to be 25kg m/s

8. Dec 14, 2013

### haruspex

What is the mass of one part?

Look at the triangle to the right of the diagram. What is the relationship between the vertical velocity before separation and the vertical velocity after separation? What in that triangle represents the change in momentum?

9. Dec 14, 2013

### Lacnez

mass of one part would be half of the original mass so 0.5 kg

is the relationship the change in vertical velocities before and after separation

and is the triangle representing the change of direction of momentum

10. Dec 14, 2013

### haruspex

Yes
What is the change in vertical velocities?
When you have answered the preceding question correctly, you will find that one part of that triangle represents the change in momentum.

11. Dec 14, 2013

### Lacnez

i believe that there is not change or very slight change in vertical velocities after the separation compared to before separation

12. Dec 14, 2013

### haruspex

Right. So in what direction is the change?

13. Dec 14, 2013

### Lacnez

the change is 7.5 degrees either direction

14. Dec 14, 2013

### haruspex

Momentum is a vector. The change in momentum is also a vector. If you subtract the old momentum vector from the new momentum vector, which way does the momentum change vector point?

15. Dec 14, 2013

### Lacnez

I just don't know,

what i'm thinking you're saying is to

original firecracker

VERTICAL 25kg m/s

HORIZONTAL 0 kg m/s

One of the pieces

VERTICAL 25 kg m/s

HORIZONTAL 3.263 kg m/s

and then subtract the components but that doesn't make sense to me

16. Dec 15, 2013

### haruspex

You're mixing up the total mass and the mass of the pieces again. I think you mean:
Indulge me... do that subtraction. What do you get?

17. Dec 15, 2013

### Lacnez

when you subtract the verticals you get 0

18. Dec 15, 2013

### Staff: Mentor

And subtracting the horizontal components (for each part!) gives....?

19. Dec 15, 2013

### Lacnez

im assuming you also get 0 for the horizontal

20. Dec 15, 2013

### Staff: Mentor

Only if you add both parts. Each part moves in horizontal direction, so it certainly has a horizontal momentum component after the split.