physics momentum question

Question

A 34-g bullet travelling at 120 m/s embeds itself in a wooden block on a smooth surface. The block then slides toward a spring and collides with it. The block compresses the spring (k = 99 N/m) a maximum of 1.2 cm. Calculate the mass of the block of wood.

Answer 1

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 = 0.5v

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.