pendulum question TPR Phys #16

[Meredith92]

I've spent so long on this question and would love some help.

This is passage #3 in TPR test 3.
There is a ballistic pendulum and it is used to calculate the velocity of a bullet from a gun.
They give us two equations:
Conservation of momentum, mv=(m+M)V
and Law of conservation of momentum: 1/2 (m+M)V^2 = (m+M)gh
where m= mass of bullet
v= muzzle velocity of bullet
V= velocity of block
h= rise in center of mass
M= mass of block and bullet

It also says energy losses due to friction in the air and the ratchet are ignored

Four trials are done with M=1kg
mass of bullet (g) height h(cm)
5 5
10 20
15 44
20 78

The question ive been stuck on asks:
If a fifth trial were attempted using a bullet of mass 30 g, the experimentalists could expect the block to swing up to a height of:
a) 132 cm
b) 156 cm
c)170 cm
d) 220 cm



I attempted this by using one of the earlier trials to solve for v where v=(( m+M) (sqrt (2gh)) ) / m

I got a v of 2000
then I plugged in a mass of 30 g and got 180cm. I dont think this approach is right though because I'm not sure I can use the same velocity since its dependent on the bullet's mass.

am I missing something really obvious here?
Thanks so much for your help!

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[cheesier]

The way I did it is by finding a trend with the given bullets velocities. Plug in their height info to find their velocities, and you'll see that the 5 g bullet had a velocity of 1, the 10 g bullet had a velocity of 2, the 15 g one had a velocity ~3, etc. You can extrapolate this for the 30 g bullet which would have a velocity ~6. Then solve 36 = 2gh and you get h~175 so I would go with c.

 

[MangoPlant]

.

 

[circulus vitios]

Doubling the bullet mass roughly quadruples the height: 5g to 10g exactly quadruples the height, 10g to 20g is a little less than quadruple. I'd guess that 15g to 30g would be a little less than quadruple, so C.

I don't really see how you could use the formulas to find an answer? Then again, it's been a long time since I've taken physics and I've yet to re-study kinematics so...

 

[dmf2682]

The way I did it is by finding a trend with the given bullets velocities. Plug in their height info to find their velocities, and you'll see that the 5 g bullet had a velocity of 1, the 10 g bullet had a velocity of 2, the 15 g one had a velocity ~3, etc. You can extrapolate this for the 30 g bullet which would have a velocity ~6. Then solve 36 = 2gh and you get h~175 so I would go with c.

This.

 

[circulus vitios]

This.

Could you do a sample calculation? I have no idea what he's doing to find the velocities.

 

[cheesier]

Could you do a sample calculation? I have no idea what he's doing to find the velocities.

I'm using the given h values and plugging them into v=sqrt(2gh)

 

[shouldvestudied]

Could you do a sample calculation? I have no idea what he's doing to find the velocities.

Given: 1/2 (m+M)V^2 = (m+M)gh

You can solve for V to get: V = sqrt(2gh)

And given: mv=(m+M)V

You can solve for v to get: v = [(m + M)*V]/m

Plug V = sqrt(2gh) into this eq to get: v = [(m + M) * (sqrrt(2gh)] / m

Then just plug in the values for each experiment to get the various v's.

 

[cheesier]

Given: 1/2 (m+M)V^2 = (m+M)gh

You can solve for V to get: V = sqrt(2gh)

And given: mv=(m+M)V

You can solve for v to get: v = [(m + M)*V]/m

Plug V = sqrt(2gh) into this eq to get: v = [(m + M) * (sqrrt(2gh)] / m

Then just plug in the values for each experiment to get the various v's.

This way you're calculating the velocity of the bullet when you could just do the velocity of the block with V = sqrt(2gh) and save yourself some time.

 

[dmf2682]

Could you do a sample calculation? I have no idea what he's doing to find the velocities.

OK sure. Make a table, the Soviet style 🙂

M. H. DeltaH. DeltadeltaH
5 5 15 9
10 20 24 10
15 44 34 ?
20 78 ? ?

Your goal is to figure out what the height at mass 30 is. The two Delta columns are changes, first change in height, then change in change in height, for a 5 gram increase. We see deltadelta is at 9, 10. Reasonable to approximate the next two numbers will be 11, 12. From these you can find the deltah's for the unknown masses, and then the heights.

I get:

M. H. DeltaH. DeltadeltaH
5 5 15 9
10 20 24 10
15 44 34 11
20 78 45 12
25 123 57
30 180

180 is ballpark for choice c.

Takes a lot longer to explain than actually do! But you can also do it graphically. The reason this approximation works is because you're increasing by constant intervals, and because essentially you're taking derivatives of the actual equations with respect to m.

Let me know if you're confused- hope it helps!