Linear momentum bullet problem

[brendan3eb]

yeah, part c was for the block :)

ahh..I thought I had the right answer originally but it didn't say if variables were supposed to be left in the problem, so I kind of doubted myself. Thanks for the help you guys.

I am learng physixguru, maybe slowly, but learning. lol

 

[physixguru]

come on buddy...these kinds of ques have been discussed a million times...

it seems u have not paid attention to them...
anywayz...

calculate the loss in kinetic energy of bullet...this equals to the work done by the block on the bullet...
then u can easily calculate the velocity of the block...using simple kinematic equations...

The third part is not understandable.:p

 

[Tedjn]

Your v_f for the bullet is correct. As physixguru says, you do not need conservation of kinetic energy to solve the other two parts. Part (c) probably means the kinetic energy gained by the block, or else (b) and (c) contradict each other.

If you want to assume that kinetic energy is conserved (i.e. the collision between the bullet and the block is elastic), you will find that v₀ does not cancel:

$$\text{From part (a), }v_{\text{block}} = \frac{v_0}{6}$$

$$\text{From the assumption that kinetic energy is conserved, }$$

$$mv_0 &=& m\cdot\left(\frac{v_0}{3}\right)^2 + 4m\cdot\left(\frac{v_0}{6}\right)^2 = \frac{mv_0^2}{9} + \frac{4mv_0^2}{36} = \frac{2mv_0^2}{9} \implies v_0^2 - \frac{9v_0}{2} = v_0\left(v_0 - \frac{9}{2}\right) = 0 \implies v_0 = 0 \text{ or } \frac{9}{2}$$

So if the given conditions are met, kinetic energy is conserved when the initial velocity of the bullet is either 0 (it isn't fired) or 9/2.

However, it is unrealistic that kinetic energy is conserved (i.e. the collision is elastic), because the impact of the bullet against the wood and the friction on the bullet when it travels through the wood will generate heat, so some of the initial kinetic energy will be lost (i.e. the collision is inelastic).

Therefore, you really cannot find out what the initial velocity v₀ is. For the given block-bullet system, there may be a particular v₀ which will cause the exit speed of the bullet to be v₀/3, but that speed is impossible to calculate and depends, among other factors, on the length of the block and the materials used.

 

[brendan3eb]

so when I get v0/6 for part A, that is precisely what they want..if so, awesome :)

then the last two parts are fairly simple