What is the speed of the bullet after it is fired into a wood block?

[FancyNut]

I'm stuck on this problem..

A 10 g bullet is fired into a 10 kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 5.0 cm across the table.

What was the speed of the bullet?

My futile attempt:

the wood block's initial momentum is zero so it's $$m_b v_b = (m_b + m_w) v_f$$

Where $$m_w$$ is the mass of the wooden block. In order to get the bullet's velocity (which is $$v_b$$) I need $$v_f$$ which is the final velocity of both the bullet and the block in this inelastic collision.

So I tried to use kinematics to get that final velocity which is equal to initial velocity from the start of motion (bullet + block) until it comes to rest after moving .05 meters.

$$v_f^2 = v_i^2 + 2 x a$$

$$0 = v_i^2 + 2 (.05) a$$

As you can see, I don't know the acceleration... so maybe I should do some force analysis? The problem didn't give the magnitude of force the bullet exerted on the block nor the coefficients for friction so I don't know...

Thanks for any help!

 

[CartoonKid]

You can use the conservation of energy here.

 

[FancyNut]

where? If I use it on the second part, where the bullet is in the block, potential energy doesn't change (distance from ground is the same) but kinetic energy is decreasing... I'm guessing it's transforming into friction/heat/other energy.


If I use it before the bullet hits the block, again, potential energy is not changing... I only have kinetic energy of the bullet just before it hits-- I don't know what to set it equal to so I can solve for velocity.

 

[CartoonKid]

You take
$$\frac{mv^2}{2}=Fd$$
F is the frictional force that stop the block after it traveled 5.0cm
Assuming all the Kinetic Energy has been changed into friction energy in the end.

 

[daveed]

whats the coefficient of friction? there is obviously friction involved, as it is wood on wood. otherwise, it would slide on forever...

 

[CartoonKid]

since you can find F, you can find coefficient of friction because $$COF=\frac{F}{Mg}$$

 

[Doc Al]

information missing

Without the coefficient of friction one cannot solve this problem.

 

[FancyNut]

You take
$$\frac{mv^2}{2}=Fd$$
F is the frictional force that stop the block after it traveled 5.0cm
Assuming all the Kinetic Energy has been changed into friction energy in the end.

How did you get that equation? I know you mean kinetic energy equals the friction force but why is the force multiplied by the distance? It makes sense if I think about it (the longer the distance the bigger the force/smaller kinetic energy gets) but I wouldn't know how to derive it myself...

 

[CartoonKid]

The F here is the frictional force. You see, the work done by frictional force is force times distance travelled. However, this question seems to have missing information after I attempted it.

 

[FancyNut]

The F here is the frictional force. You see, the work done by frictional force is force times distance travelled. However, this question seems to have missing information after I attempted it.

Hmm I skimmed through my text and found a chart for some friction constants. I tried your formula then and it worked... weird considering just about every other problem that had friction in it, I was given the coefficients... this is www.masteringphysics.com btw.

I think there's another way though. If your formula relied on work (force * distance) then there must be because this is part of HW 9 which just covers momentum/beginning of energy.

 

[Andrew Mason]

This problem was discussed last week at: https://www.physicsforums.com/showthread.php?t=50979