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phys_08 · Jan 2, 2010, 07:17 PM

A 13-g bullet traveling 230m/s penetrates a 2.0Kg block of wood and emerges going 170m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

galactus · Jan 3, 2010, 01:43 PM

Originally Posted by phys_08

A 13-g bullet traveling 230m/s penetrates a 2.0Kg block of wood and emerges going 170m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

I am a little rusty on my elastic and inelastic collisions, but one would think to use the formula:

$\frac{1}{2}m_{1}v_{1i}^{2}+\frac{1}{2}m_{2}v_{2i}^ {2}=\frac{1}{2}m_{1}v_{1f}^{2}+\frac{1}{2}m_{2}v_{ 2f}^{2}$

This looks nasty, but really just involves the initial and final velocities of the block and bullet.

The initial velocity of the bullet is 230 m/s

The final velocity of the bullet is 170 m/s

The initial velocity of the block is 0 m/s

The final velocity is what is asked for.

$\frac{1}{2}(2)(0)^{2}+\frac{1}{2}(.013)(230)^{2}<br /> =\frac{1}{2}(2)v^{2}+\frac{1}{2}(.013)(170)^{2}$

Solve for v.

This exercise is an example of how kinetic energy works. We always here about the horrible consequences of an asteroid impact. The Earth is so huge compared to an asteroid that may only be a few miles across. The thing is the velocity at which it is traveling is so fast that it makes all the difference. Suppose the mass of the bullet in this exercise was kept the same, but its velocity raised to, say, 5000 m/s. It would do a lot more damage to the block. This is the idea in high-tech weaponry. If a projectile is going fast enough, one does not even need explosives to cause damage.

InfoJunkie4Life · Jan 4, 2010, 11:03 AM

The wood slowed down the bullet by absorbing some of the energy. Calculate the amount of energy in the bullet prior and after the wood collision and subtract them.

$E = MV$

$E_2 - E_1 = E_{absorbed}$

Then you can take that answer and plug it into the first equation to find out how fast the wood will be moving by solving for velocity. Make sure you do it all in Kilos.

$M_1{V_2} - M_1{V_1} = M_2{V_3}$

Solve for $V_3$ .

$\frac{M_1|{V_2} - {V_1}|}{M_2} = V_3$

The absolute value is just to prevent a negative velocity since no vectors are involved.

Unknown008 · Jan 6, 2010, 10:09 AM

Yeah, I'm a bit confused here. I admit that both methods should work, but I'm having two different answers. I get about 12 m/s using law of conservation of energy and 0.3 m/s using principle of conservation of momentum :confused:

The 12 m/s seems the more realistic though...

EDIT: Oh, and infoJunkie, energy is not given by mv, but that is momentum. Energy is (mv^2)/2

Unknown008 · Jan 7, 2010, 12:00 PM

Oh, perhaps it's an inelastic collision and that explains why the principle of conservation of momentum is not valid. :confused:

InfoJunkie4Life · Jan 7, 2010, 03:04 PM

You can look at it as 2 systems involving force and mass.

$F = ma$

The first stystem, the bullet has a force exerted upon it by the block of wood, this force causes a negative change in velocity. Any change in speed or direction is acceleration.

$F = m(v_2 - v_1)$

When force is applied, work is done (energy is released), in this case in the form of kinetic energy. Thus the Force is conserved, and applied to the wood that absorbed it. So now we get

$m_1(v_2 - v_1) = F_1 = F_2 = m_2(v_4 - v_3)$

I know its not energy, I was minorly confused, I know conservation of momentum.

I don't quite understand, why they result in different amounts. I'll look into it...

galactus · Jan 7, 2010, 03:52 PM

The correct method is

$(.013)(230)=2v+(.013)(170)$

Solving for v gives us

v=.39 m/s

I had it set up right, but should not have squared the velocities.

You all were correct. I made a booboo. I used energy when it should have been momentum.

Unknown008 · Jan 7, 2010, 08:40 PM

Perhaps it's because it's an inelastic collision, only momentum is conserved and energy is 'lost' to surroundings.

galactus · Jan 8, 2010, 05:02 AM

Yes, that's why.

InfoJunkie4Life · Jan 8, 2010, 01:02 PM

How can you tell. There is no mention of lost energy, nor does the equation recognize it. I don't get quite how momentum and kinetic energy are calculated differently in this situation.

Unknown008 · Jan 8, 2010, 10:29 PM

That's a situation that you have to know, like in the case where two projectiles stick together after collision, that is a case of 'perfectly inelastic' collision so my teacher calls it. The simple fact that the bullet got through the wooden block shows that it's an inelastic collision.

If that was to be an elastic collision, the bullet would have been going in the opposite direction or come to rest because the other body we are dealing with is about 150 times more massive.

This has helped me recall some things that I've forgotten. Thanks galactus you helped me in a certain way through your mistake :)