A 13 g bullet traveling 213 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 152 m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

Question

roadjester · Answer

Since this is penetration rather than collision, I'm not TOO sure, but perhaps you could modify the Conservation of Linear Momentum and Conservation of Energy?

anonymous · Answer

KE = 1/2 mv^2
The bullets kinetic energy before impact:
KE = 1/2 (0.013 kg)( 213 m/s)^2
KE of bullet = 295 Joules

Now the bullet enters the 2kg block and loses some of that KE to the block. But first have to find out how much KE the bullet lost when traveling thru block and exiting at 152 m/s:

KE = 1/2 (0.013 kg) (152 m/s)^2
KE of bullet after passing thru block = 150 Joules

So the bullet lost this KE and transferred it to the block, 
 295 J - 150 J = 145 Joules

So now just have to calculate how much velocity the block gained from that 145 Joules of energy transferred:

145 J = 1/2 ( 2 kg) ( v )^2
Solve for V:
145 = v^2
v = sq rt (145)

answer: Velocity of block = 12 m/s

anonymous · Answer

@biire2U, that isn't correct. We haven't gone over anything with Joules, so I don't think that would be the way to solve it. It's an online program, so it just automatically grades the answer I submit, and that answer is not 12 m/s :/

roadjester · Answer

Have you gone over linear momentum? and how many attempts do you get?

anonymous · Answer

That's what the lesson is now, but I've been on homebound so I haven't quite gotten it. I think I have 3 more attempts left.

roadjester · Answer

If we focus purely on linear momentum, then it looks like an inelastic collision/penetration although I'm not sure how to go about that...

whpalmer4 · Answer

The amount of momentum transferred is the amount lost by the bullet, right?

$$\Delta p_{bullet} = m_{bullet}*v_i - m_{bullet}*v_f$$

roadjester · Answer

@whpalmer4 this is purely me, but could you enlarge the text? I'm having a bit of trouble seeing it.

whpalmer4 · Answer

the block collects the same amount of momentum, so we should be able to find its velocity after the penetration by $$\large \Delta p_{block} = -\Delta p_{bullet} = -m_{block}*(0-v_{block})$$

previous was: $$\large \Delta p_{bullet} = m_{bullet}*v_i - m_{bullet}*v_f$$

whpalmer4 · Answer

Agree?  Disagree?

roadjester · Answer

I agree with the momentum of the bullet, trying to comprehend what you're doing with the line involving the block

whpalmer4 · Answer

If I plug in the numbers, I get the bullet decreasing its momentum by 0.793 and the block gets a velocity of 0.4 m/s, which seems more reasonable.

whpalmer4 · Answer

momentum is conserved, so the block goes from 0 to its final speed with the momentum the bullet lost.  p = mv  if I'm remembering things correctly

anonymous · Answer

.4 was correct!! Thank you very much, you explained it wonderfully.

roadjester · Answer

Hmm ok, if @heylainey get's it then I'll just think about it myself.
I think  I need to refresh my mechanics.

whpalmer4 · Answer

sometimes an old dog still remembers his tricks :-)

roadjester · Answer

lol

whpalmer4 · Answer

so the bullet lost momentum, giving it a negative value of delta p (change in momentum) by $\Delta p = m \Delta v = 0.013(152-213)$.  the block gained momentum, so it has a positive value of $\Delta p = 2(0-v_f)$.  And because no momentum was lost, the two of them together must sum to 0.  the change in momentum of the block is going from an initial velocity of 0 to the final velocity, so that gives us our formula

$$0.013(152-213) = 2(0-v_f)$$$$\frac{0.013(152-213)}{-2} = v_f \approx 0.4$$

anonymous · Answer

Very good whpalmer4, I should of noticed that 12 m/s was way to fast for a 2kg block to move and tried your approach with momentum instead of KE

whpalmer4 · Answer

Certainly wouldn't want to be on the receiving end of a bullet whose impact would produce that sort of acceleration of a 2 kg block :-)

whpalmer4 · Answer

A .50 caliber round from a sniper rifle, weighing in at 42.8 grams (think golf ball weight), leaves the barrel at about 900 m/s.  If that hit the same block and slowed down to the same exit velocity, that'd make the block move at about 16 m/s.  So not impossible for a bullet to achieve that sort of acceleration of a 2 kg block (assuming I didn't make any errors), but I'm sticking by my statement that you don't want to get hit by such a bullet!  That same round can penetrate 2 cm of steel armor plate at 600 meters range, so it'd hardly even slow down going through you...