# Homework Help: Impulse/Momentum Question

1. Jul 27, 2010

### coffeem

A bullet of mass 60g is fired into a 20kg block of wood which is suspended on a light inextensible wire to form a ballistic pendulum. After penetration the bullet remains in the wood. The vertical position of the wood-bullet combination increases by 28cm. Calculate the velocity of the bullet?

I know that you will have to use conservation of momentum here... However I am unsure of how i am going to do this with the given information. Ant tips? thanks

2. Jul 27, 2010

### Staff: Mentor

Yes, you'll need conservation of momentum--among other things.

Think of this as having two steps:

(1) The bullet hits the wood. How can you relate the speed of the bullet before the collision to the speed of the 'bullet + block' immediately after the collision?

(2) The block rises after the collision. How does the height relate to the initial speed of the block?

You might have to work backwards.

3. Jul 27, 2010

### coffeem

Thanks. I have tried thinking about what you have said, but am still coming up very very short.

1) I think that I would have to look into conservation of energy. In which case I would assume that the energy of the bullet alone would have to have the same energy as the wood and the bullet combines (given that the block is stationary it will have no kinetic energy). However I cannot see how this helps me???

2) I am really gussing here, but I am going to assume that it would be possible to work out the speed of the block as some sort of projectile? With only gravity resisting the motion??? I really do now know where I need to go with this...

If you could give me some more advice I would appreciate it. Thanks

4. Jul 27, 2010

### merryjman

Your guess to 1) is not correct. The bullet has to do some damage to the block in order to become embedded in it. This converts some energy into thermal energy, which means you can't use conservation of energy in that scenario. What do you do, then? If you knew the speed of bullet+block after collision, it would be easy, so as Doc said you need to work backwards.

Remember this is called the ballistic pendulum and the word "pendulum" is meant to give you a clue about what is going on after the bullet gets embedded in the block.

5. Jul 27, 2010

### Staff: Mentor

You cannot assume that energy is conserved during the collision. The collision is completely inelastic--the bullet becomes embedded in the block. Hint: This is where momentum conservation comes in.

What's conserved as the block swings up?

6. Jul 27, 2010

### coffeem

OK I have shown I know very little so far... So here goes for my next attempts at this problem.

If I work out the velocity of the block + bullus using: v^2 = u^2 + 2as

given that u = o then v = root(2as) = 2.34m/s

Then I can use conservation of momentum to say: that the momentum of the bullet must be equal to the momentum of the bullet and the block.

So: 60ee-3*v = 20.06*2.34

giving the velocity of the bullet as: 782m/s

am i anything like right?? thanks

7. Jul 27, 2010

### Staff: Mentor

This is incorrect. That equation is only good for uniformly accelerated motion, which is not the case here. (The rising block is not a free projectile.) Edit: Please ignore this comment. You are completely correct!

Again: What's conserved as the block rises?

Exactly right. First fix up the previous part and then redo this.

Last edited: Jul 27, 2010
8. Jul 27, 2010

### coffeem

When the block rises would the kinetic energy be conserved with the potential energy?

in which case: mgh = 1/2mv^2

However this would give me the same result so has to be wrong....

9. Jul 27, 2010

### Staff: Mentor

Correct!

Oops... I must have been asleep at the wheel! That part was correct! (Sorry about that! :uhh:)

Let me check over the rest of it.

10. Jul 27, 2010

### Staff: Mentor

You are perfectly correct!

Just be careful that you are using v^2 = u^2 + 2as for the right reasons. The rising block is not a projectile--the acceleration is not equal to g. But it is true, per conservation of energy, that:
1/2mv^2 = 1/2mu^2 + mgh

v^2 = u^2 + 2gh

Last edited: Jul 27, 2010
11. Jul 27, 2010

### mizzy

This is an example of an inelastic collision. First of all you should know that momentum is conserved. Knowing that you should solve that. Find the momentum before and let it equal to the momentum after.

12. Jul 27, 2010

### mizzy

ooops...i didn't see the previous post...hehe.

13. Jul 27, 2010

### coffeem

Thanks for all of the help...

However I am going to be cheeky now and ask for some help on another questions.

A cork ball of mass 100g at rest is released from a height of 8.5m about the launch of a 100g dart which is projected vertically upwards with an initial velocity of 12m/s. Upon impact the cork and the dart stick together. Calculate the velocity immediately after the collision.

My first thoughts on this are. I have to work out when the two items collide. I have to work out the velocities of the items before the collisions. I would then be able to use conservation of moment to work out the final velocity? Idea ideas of how else I should go about this?

14. Jul 27, 2010

### Staff: Mentor

Sounds good to me.
The first part is a projectile motion exercise. Write an expression for the position of each as a function of time. Then you can solve for the point where they meet.

15. Aug 11, 2010

### coffeem

Hi, I am coming back to this problem. And am still stuck on it...

I have two sets of suvat terms like i would if we were dealing with a projectile motion. For the cork I have:

s = ?
u = 0
v = ?
a = g
t = ?

for the dart i have:

s = ?
u = 12
v = ?
a = -g
t = ?

However I do not know where to go from here? Any ideas thanks.

16. Aug 11, 2010

### Staff: Mentor

You're on the right track.

The acceleration of both cork and dart is -g. You might want to use a more complete version of 'SUVAT', such as:

$$x = x_0 + v_0 t + (1/2) a t^2$$

I suggest that you measure the position from the ground. What's the initial position of the cork?

17. Aug 11, 2010

### coffeem

How is the cork -g? If it is falling downwards surely it must be +g?

18. Aug 11, 2010

### Staff: Mentor

I suggest using a consistent sign convention. Up is positive and down is negative. The acceleration of both objects is -g (in that convention).

You could use a different sign convention for each, but that makes things harder to keep track of, in my opinion.