# Momentum problem

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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haruspex
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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.

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

mfb
Mentor
I tried out the question but I don't get far cause I really have no clue on how to do this question.
Please show your work then, so we can see where you got stuck.

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.

mfb
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.

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
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...

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

haruspex
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Gold Member
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
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?

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

haruspex
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Gold Member
mass of one part would be half of the original mass so 0.5 kg
Yes
is the relationship the change in vertical velocities before and after separation
What is the change in vertical velocities?
and is the triangle representing the change of direction of momentum
When you have answered the preceding question correctly, you will find that one part of that triangle represents the change in momentum.

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

haruspex
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Gold Member
i believe that there is not change or very slight change in vertical velocities after the separation compared to before separation
Right. So in what direction is the change?

the change is 7.5 degrees either direction

haruspex
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Gold Member
the change is 7.5 degrees either direction
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?

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

haruspex
Homework Helper
Gold Member
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
You're mixing up the total mass and the mass of the pieces again. I think you mean:
each piece of original firecracker VERTICAL 12.5kg m/s HORIZONTAL 0 kg m/s
One of the pieces later, VERTICAL 12.5 kg m/s HORIZONTAL 1.63 kg m/s
Indulge me... do that subtraction. What do you get?

when you subtract the verticals you get 0

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

im assuming you also get 0 for the horizontal

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

would it after the split subtracting the horizontal vectors give you some were around 6.52k kg m/s

mfb
Mentor
No. It would help if you write more clearly what you did where, what you got and so on. Posting some numbers makes it hard to guess what you did.

i took 25sin7.5 got 3.263 the subtracted

3.263 - (-3.263) and got 6.52 as an overall horizontal

haruspex
Homework Helper
Gold Member
i took 25sin7.5 got 3.263 the subtracted

3.263 - (-3.263) and got 6.52 as an overall horizontal
Two things wrong with that:
- each piece is 0.5kg, not 1kg
- we want the change in momentum of one piece; what was its horizontal momentum before the split and what is it after the split?

The horizontal before the split was 0 cause it was going vertical

For one of the pieces I did now

P=mv
(0.5)(25)
=12.5

Then
12.5sin7.5

I get as a horizontal 1.6315

Then subtract

1.6315-(-1.6315) = 3.263