1. Apr 13, 2005

### NoHeart

I have three questions that I'm having a hard time with. Any help would be greatly appreciated...

easiest one first:

1. A 3kg ball of putty moving at 1m/s collides and sticks to a 2kg ball at rest. The putty and the ball then move with a momentum of what?

so far i have initial momentum of putty at 3Ns (3kg*1m/s) and then i'm not quite sure... initial momentum of other ball is zero? (2kg*0m/s=0Ns)
so do they both keep moving with a momentum of 3Ns? or is the force not enough for them to move, and it's 0Ns? i'm thinking the momentum cannot remain constant because the mass has changed...so would it be correct to say 3kg-2kg=1kg, and that momentum would equal 1Ns? i'm thinking this is the answer (1Ns) but i'm not sure if i'm grasping the concept correctly...

2. a 20g bullet travels at 350m/s, strikes metal plate at an angle of 30, bullet ricochets off at same angle with a speed of 320m/s- What is the magnitude of the Impulse the wall gives to the bullet?

this is giving me a lot of trouble, i'm not sure what to do with the angles...the diagram shown has the bullet traveling at an angle pointing up towards the wall, and ricocheting off upwards, so the angles are in reference to the y-axis...
if impulse is a change in momentum, i have initial momentum at 7Ns (20g*350m/s) and final momentum at 6.4Ns, so the impulse would be
6.4-7= -0.6Ns
that's not anything close to what i think the answer is, i know i should be using sin or cos30 at some point, but where? and is the initial momentum going to be negative because it is coming from the negative x-axis?
i am so confused...

3. a 10.5g bullet strikes a 3kg wood pendulum and imbeds in the wood. the pendulum rises by 0.22m after impact, What was the initial velocity of the bullet?

okay, so lost. the initial momentum has to equal the final momentum, right? but how do i find momentum without velocity? i could find the final momentum using the distance of 0.22m, but i don't know the time. where do i even start here?

2. Apr 13, 2005

### learningphysics

Yes. Momentum is conserved. They keep moving with 3Ns.

Suppose the initial momentum is vector a. The final momentum is vector b. We want the magnitude of the vector b-a (use the parallelogram rule for adding vectors...). Draw a sketch of the vector b-a (sketch b... add -a). You'll need to use a little geometry to get the magnitude of b-a.

There are 2 stages to this problem. First there is the bullet striking and embedding itself in the wood... here momentum is conserved (momentum is in the horizontal direction and there are no forces acting in the horizontal direction).

In the second stage(pendulum moving upwards with the bullet) momentum is not conserved because we have an external force (gravity) acting against the motion of the pendulum. But we have energy conservation here.

Try working backwards from stage 2 to stage 1.

3. Apr 13, 2005

### NoHeart

so for the 3rd one, i need to deal with energy...kinetic and potential energy are conserved...? so i can say kinetic energy after the collision is1/2mv^2 and that is equal to the potential energy of the pendulum after impact, which is mgh...so m=10.5g+3kg=3.0105 and h=0.22m
this leads to final velocity at 2.08m/s
final momentum is 0.0105kg*2.08m/s+3kg*2.08m/s=6.26m/s
how do i use final momentum to get initial velocity if momentum is not conserved? are you sure the initial momentum isn't equal to the final momentum?

4. Apr 14, 2005

### learningphysics

Good work!

Yes, you are right momentum is conserved (this is stage 1 now - the collision). You can calculate initial velocity just like you said.

What I meant was that you can't use conservation of momentum in stage 2 (when the pendulum goes up).

5. Apr 14, 2005

### NoHeart

thank you and goodnight

thank you so much! i ended up with the answer in terms of acceleration though, 596.2m^2/s
i think that'll work, the exact question was "how fast was the bullet moving" velocity wasn't specifically mentioned.
now i'm gonna go study vectors...

6. Apr 14, 2005

### learningphysics

Careful with your units. I noticed that you wrote 6.26 m/s instead of 6.26kg*m/s for your momentum.

When you set 0.0105 kg * v = 6.26 kg*m/s

v comes out to 596.2 m/s (units come out to m/s)

Also, another thing... try not to round much until you need your final answer... I got 595.4 m/s when I didn't round till the end.

Good luck.

7. Apr 14, 2005

### NoHeart

thanks again,LP, that certainly makes a difference in my answer...

for #2, in my AC electronics class a couple semesters ago we learned how to convert polar coordinates to rectangular- can i apply this to the problem? if i'm starting with the two polar coordinates of 350m/s at an angle of 30(or is it 150?) and 320m/s at an angle of 30, should i convert them to rectangular and add them? and is that answer the magnitude?

8. Apr 14, 2005

### learningphysics

Hi NoHeart. Yes, you can certainly do it this way. I'd sketch in the vertical and horizontal components of each vector. This way you can see the angles, and you can get the vertical and horizontal component by using the sine or cosine. You just have two right triangles.

Essentially you are getting the horizontal and vertical components of the final momentum... and also the horizontal and vertical components of the initial momentum.

Then the horizontal component of the impulse is the horizontal component of the final momentum minus the horizontal component of the initial momentum. And the vertical component of the impulse is the vertical component of the final momentum minus the vertical component of the initial momentum.

If the initial velocity is upwards to the right (wall is on the right side) making an angle of 30 degrees with the vertical... then the horizontal component of momentum is mvsin(30) and the vertical component is mvcos(30). Do you see this with the right triangle? Be careful regarding signs.

When you get the horizontal and vertical components of the impulse use the pythagorean theorem to get the magnitude.

Hope this helps.

9. Apr 14, 2005

### the_d

For question number 1, how would you find the velocity of the two bodies after collision?

10. Apr 14, 2005

### learningphysics

We have the momentum = 3Ns or 3kg*m/s

The total mass of the ball and putty=5 kg.

Momentum = m v

So v=momentum/m = 3/5 = 0.6 m/s

11. Apr 15, 2005

### NoHeart

still have a problem

my way i get .59 at an angle of -30
this is a multiple choice question and my options are 300 Ns, .30Ns, .52Ns, and 6.7Ns

the way i got .59 was
a=mvsin30=3.5
b=mvsin30=3.2

a=mvcos30=6.06
b=mvcos30=5.54

Impulse a= 3.2-3.5=-.3
Impulse b= 5.54-6.06= -.52

Magnitude^2= -.3^2+-.52^2
Magnitude=.59

what am i doing wrong?

12. Apr 15, 2005

### learningphysics

Hi Noheart. The initial horizontal momentum is 3.5 Ns. The final horizontal momentum is -3.2Ns (since it's in the opposite direction). So the horizontal component is -3.2-3.5=-6.7

The vertical impulse is -0.52 as you calculated.

magnitude^2 = (-6.7)^2+(-0.52)^2
magnitude = 6.72

13. Apr 15, 2005

### NoHeart

will you marry me?
just kidding...now i see what you meant about being careful of signs...i am so happy that i actually understand this now!

14. Apr 15, 2005

### learningphysics

:grumpy: Why aren't the marriage proposals ever serious?! :rofl: