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[SD-Ness]

I'm taking AP Physics C: Mechanics in high school. My teacher assigned us these problems, but we've never gone over how to do them. I've read all the sections in the textbook and I'm still not getting some of them so if anyone can give me a hand here that'd be awesome.

Chapter 8: Systems of Particles and Conservation of Linear Motion

Impulse and Average Force

61. A meteorite of mass 30.8 tonne (1 tonne = 1000 kg) is exhibited in the American Museum of Natural History in New York City. Suppose that the kinetic energy of the meteorite as it hit the ground was 617 MJ. Find the impulse I experienced by the meteorite up to the time its kinetic energy was halved (which took about t = 3.0 s). Find also the average force F exerted on the meteorite during this time interval.

I have already found the impulse.

vf = vi + at
0 = vi + (9.8)6
vi = 58.8

pf - pf = i
0 - (58.8)(30.8) = i
i = 1.81

How do I find the average force though?

i / t = 1.81 / ?

The answer I get when plugging in 3 for the ?, is not the answer in the back of the book.

71. A proton of mass m undergoes a head-on elastic collision with a stationary carbon nucleus of mass 12m. The speed of the proton is 300 m/s. Find the velocity of the proton after the collision

Having trouble here...

Perfectly Inelastic Collisions and the Basllistic Pendulum

84. Tarzan is in the path of a pack of stampeding elephants when Jane swings in to the rescue on a rope vine, hauling him off to safety. The length of the vine is 25 m, and Jane starts her swing with the rope horizontal. If Jane's mass is 54 kg, and Tarzan's is 82 kg, to what height above the ground will the pair swing after she grabs him?

So I've gotten the Jane's velocity:

pe = ke
mgh = 1/2mv^2
54(9.8)25 = 1/2(54)v^2
v = 22.14

I then plugged this into:

P + P = P
mv + mv = mv
(54)(22.14) + (82)(0) = 136V

But there's nothing I can do with this to find their height...



Thanks.

 

[nitewulf]

problem 71 is a conservation of momentum problem in 1 dimension, assuming perfectly elastic collision and negligible frictional effects,

m1v1 + m2v2 = m1v1' + m2v2' = total momentum

where the total momentum of the system is just (300m/s*m1).
what we are interested in is v1', the velocity of the proton after the collision.

now from newton's energy conservation laws, we also know that the kinetic energies of the system is conserved.

so we can write down,

.5m1v1^2 + .5m2v2^2 = .5m1v1'^2 + .5m2v2'^2

after some algebraic manipulations and solving the above equation for v1', you should end up with,

v1' = [(m1-m2)/(m1+m2)]*v1 + [2m2/(m1 + m2)]*v2

notice the second part on the right side of the equation drops off as v2=0.

so plugging in the values, you should end up with

v1' = -253.85 m/s

edit: sorry, miscalulated and went "huh" while looking at the results.

 

[Dilbert]

You guys rock.