# Using elastic equations to calculate height

1. Jul 30, 2013

1. The problem statement, all variables and given/known data
2. A 0.30 kg bullet is fired from a gun with a velocity of 410 m/s. The bullet then becomes embedded into a piece of wood with mass 5.5 kg hanging as a pendulum at rest. To what height will the piece of wood swing?
m1=.3kg
v1=410m/s
m2=5.5kg
v2=0

2. Relevant equations
m1v1 + m2v2 = m1(v1’+ v2’)

3. The attempt at a solution
i plugged in the variables (.3*410+5.5*0=.3(v1'+v2')), but it comes out to 410. I know that in a perfect inelastic equation all energy should transfer, but how can i use this to calculate the height the pendulum will reach?

2. Jul 30, 2013

### technician

Do you know what happens to momentum when the bullet collides with the block?
In an in elastic collision kinetic energy is not conserved.

Last edited: Jul 30, 2013
3. Jul 30, 2013

### haruspex

No, that's for a perfectly elastic collision. You have successfully calculated the velocity immediately after the collision. From that point, you can take work as being conserved. How much work energy does the system have?

4. Jul 30, 2013

### technician

I do not think that the velocity after the collision has been calculated!!
It should be approx 20 m/ s
There is also something unrealistic about these numbers....a bullet of 300 g travelling at 410m/ s ??

It is a cannon!

Last edited: Jul 30, 2013
5. Jul 31, 2013

So I use K=1/2mv^2, when m=5.8 (mass of bullet + mass of wooden block) and v=20.8 (final velocity)?

Last edited: Jul 31, 2013
6. Jul 31, 2013

### technician

That is the correct way to calculate the KE after the collision.
The numbers do not look realistic to me !!
Have you checked that the bullet is indeed 300 g and the speed 410 m/ s ( much greater than the speed of sound!!)

7. Jul 31, 2013

I did check and they are the correct numbers. How would I solve for final velocity? I do not get the correct answer (21), and I don't understand why.

8. Jul 31, 2013

Which is the correct equation for inelastic collisions?
m1*v1+m2*v2=m1(v1'+v2')
or
m1*v1+m2*v2=(m1+m2)vf

9. Jul 31, 2013

### technician

For the collision
m1 x v1 = (m1+m2) x v2
This gives v2 = 21.2 m/s

Where did the question come from? The numbers relate more to a cannon than a rifle/handgun.
You will find that you get a totally unrealistic final answer for the final height of the block!!

PS....you were not asked to calculate it but your numbers have a 'bullet' with kinetic energy of just over 25000 (25kJ !!) Joules

Last edited: Jul 31, 2013
10. Jul 31, 2013

the question was made by my teacher. He also gave me the other equation i posted which is why i could not solve for the velocity.

also, I noticed the kinetic energy but he makes a lot of problems with numbers similar to these so i thought nothing of it.

11. Jul 31, 2013

so after solving for the final velocity what equation do I use to solve for the height?

12. Jul 31, 2013

### technician

Carry on with it....you will learn the correct method and equations but it is a great pity that the answer you get will be.........ridiculous!...sorry about that.
Make certain you understand the physics and don't worry too much about the reality.
If the mass of the bullet was 0.03kg it would be more realistic.

PS. KE to PE ?

13. Jul 31, 2013

I see Ke=Pe so when you solve you will be left with height. Thanks for this i did not understand it when my teacher taught this

14. Jul 31, 2013

### technician

Work through it and post your answer, I will tell you what I get and then we can laugh about it!!

15. Jul 31, 2013

KE= 1/2mv^2
KE=1/2*5.8*21.2^2
KE=1303
KE=PE
1303=mgh
1303=5.8*9.8*h
h=22.9

16. Jul 31, 2013

### technician

That is correct, the calculation shows that the block will swing to a height of 23m....this is about 7 stories !!!!
Do you see how unrealistic the question is.
Here are some more interesting facts that come from the numbers in your question
1) a bullet fired vertically upwards with a velocity of 410m/s would reach a height of 8400m (25,000 feet !!)
2) The KE of the bullet is 25,215J, the KE of the block is 1,280J. This means that 23,935 Joules of energy were deposited in the block...it would be blown to smithereens.
It will be a good exercise for you to confirm these numbers.

If the mass of the bullet had been 0.03kg could you redo the calculations and see what you would get.... much more believable.
anyway...you got the correct (ridiculous) answer.

17. Jul 31, 2013