Stanton D. answered • 3d

Tutor to Pique Your Sciences Interest

So Alex,

This both IS and IS NOT a momentum question!

Why both? You can certainly see that momentum is conserved when the bullet hits the block -- it ALWAYS is. So the momentum of the combined bullet+block is just the same as that of the bullet initially, namely 34g * 120 m s^-1 = 4.080 kg m s^-1.

BUT, all you can tell from the compression of the spring is the **kinetic energy** it stored, namely, (1/2) kx^2, or 0.5*99 N m^-1 * (1.2 cm)^2 = 0.007128 kg m s^-2 m^-1 m^2 = 0.007128 kg m^2 s^-2

Now, what is the relationship of those momentum and kinetic energy expressions?

P = *mv* ; KE = (0.5)*mv*^2

So,__ for the combined bullet+block__, as it hits the spring (before starting to compress it!), KE/P = 0.5*v*

That is: 0.007128 kg m^2 s^-2 / (4.080 kg m s^-1)*0.5 = 0.0008725 m s^-1 = *v *(That's hardly moving!)

Since that had momentum of 4.080 kg m s^-1, you can figure *m* : = *mv/v *= 4.080/0.0008725 kg = 4676 kg

The initial bullet doesn't even register in that mass (but, if the combined mass were say 100 g then you would need to subtract it out).

Pretty hefty block of wood, 4 metric tonnes ....

-- Cheers, --Mr. d.

Grigoriy S.

2d