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# Help with physics!!!!!!!

Discussion in 'Off-Topic' started by eddy, Mar 21, 2004.

1. ### eddyBanned

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Help with physics!!!!!!!

Well i have this phsyics homework question and i don't understand at all how to solve it, any help/hints would more then appreciated. Thanks in advance.

A 2.0-kg bag is held by a string to the ceiling as shown in the diagram below. A 10-g bulled travelling at 200 m/s strikes the stationary bag. The height of the bag after the collision is 10.0 cm. Assuming there is no friction, determine the speed, in mtres per second, of the bag after the collision.

2. ### dantoseIncGamers Member

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conservation of momentum. mass of bullet*velocity of bullet=mass of bag+bullet*velocity after impact

3. ### eddyBanned

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for the second half is it mass of bag + mass of bullet? just a little verification.

4. ### tintrailIncGamers Member

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i would say yes.. i would need a physics person to back me up though, actually i am 99 percent sure that it is indeed mass of bag + mass of bullet

5. ### Killswitch EngageIncGamers Member

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Yes, you are correct sir.

V = Speed of Bullet
U = Speed of Bullet and Bad after collision
m = Mass of Bullet
M = Mass ob Bag

By the conservation of mumentom (sp? have no idea how to spell it) law (rule, whatever), you get this equation:

m * v = (M+m) * u

Now let's see what we have

V = 200 m/s
M = 2 kg
m = 0.01 kg

Just take those into the quation and it's very simple afterwards:

0.01 * 200 = (2 + 0.01) * U

2 = 2.01 * U

U = 0.995 meters per sec (m/s).

Is that the answer in the book?

6. ### UnderseerIncGamers Member

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Actually, the best bet is usually to go off of energy.

We know that the bullet and the bag eventually reached a height of 10cm (or 0.1m). At that height it had zero velocity, so it was all potential energy and no kinetic energy.

Immediately after the impact, it was at the lowest point, so the energy was all kinetic and no potential.

The energy after and the energy before should be more or less equal (not counting non-conservative forces from friction with the air, friction in whatever string/chain is holding the bag, etc.). Using Killswitch's notation:

Energy before = Energy after
(1/2)(m+M)u^2 = (m+M)gh

m+M is on both sides, so we can simplify to

u^2 = 2gh
u = Â± sqrt(2gh)

Plugging in numbers we get

u = Â± sqrt[ 2 * (9.8 m/s/s) * (0.1)m ]
u = Â± 1.4 m/s

We will assume that the coordinate system is such that the velocity of the bag and bullet were positive immediately after impact, so we can ditch the plus or minus sign and assume positive. (Note: if the diagram you didn't show us includes a coordinate system, use that to choose plus or minus.)

If you work it backwards, you'll see that Killswitch's velocity wouldn't have reached 10cm.