What Force Does the Wood Exert on the Bullet?

  • Thread starter Thread starter mrnastytime
  • Start date Start date
  • Tags Tags
    Dynamics Force
AI Thread Summary
The discussion focuses on calculating the force exerted by wood on a bullet that embeds itself within it. The bullet, with a mass of 0.002 kg and an initial velocity of 510 m/s, travels 0.78 m before stopping. Using the equation for acceleration, the calculated acceleration is approximately -166730.76 m/s², leading to a force of -333.46 N when applying F=ma. Participants suggest checking for unit inclusion, adjusting significant figures, and considering potential errors in the online homework system. Ultimately, the correct answer for the force is 333 N.
mrnastytime
Messages
28
Reaction score
0

Homework Statement


A bullet of mass 0.002 kg initially moving at 510 m/s embeds itself in a large fixed piece of wood and travels 0.78 m before coming to rest. Assume that the acceleration of the bullet is constant.


Homework Equations


What force is exerted by the wood on the bullet?


The Attempt at a Solution


In order to find the force, I had to solve for acceleration using the velocity vs function equation: v^2=V0^2 + 2a\Deltax
a=v^2-V0^2/2\Deltax
a=0-(510)^2/2(.78)
I plugged in the numbers to the equation f=ma, but it doesn't work. Can anybody help me?
 
Physics news on Phys.org
It seems to work fine using the equations you posted, which seem correct. Can you show your work, and explain what you mean by the statement, "It doesn't work"?
 
a=0-(510)^2/2(.78)
a=-166730.76
f=0.002kg * -166730 m/s^2
f=-333.46

I entered this answer into my homework online and it is incorrect
 
Nice work; your answer is correct. Should you include units when you enter an answer into your homework on-line? If so, add capital N, for Newtons. If that fails, try removing the negative sign (even though the negative sign is correct). If that fails, try entering your answer to only three significant digits, instead of five. If that fails, the on-line program might be wrong or malfunctioning.
 
Last edited:
Thanks! the answer is 333 N
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top