A 23-g bullet traveling 260 m/s penetrates a 5.0 kg block of wood and emerges cleanly at 215 m/s

Question

A 23-gg bullet traveling 260 m/sm/s penetrates a 5.0 kgkg block of wood and emerges cleanly at 215 m/s If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

Express your answer using two significant figures.

Answer 1

Let

m = 23 g be the mass of the bullet,

M = 5 kg = 5000 g be the mass of the block,

v₀ = 260 m/s be the initial speed of the bullet,

v₁ = 215 m/s be the final speed of the bullet,

V be the final speed of the block (to be calculated).

With these kinds of problems we need to decide what is conserved in the interaction --

energy, or momentum, or something else.

For that we need to look for clues in the statement of the problem.

I do not see anything in the statement that would suggest conservation

of energy; in fact it would be surprising given energy loss due to friction

of the bullet inside the block.

But the statement of the problem contains the word "cleanly".

I believe that it means that there are no wood chips flying out.

So the bullet and the block are the only masses involved, and we can apply

conservation of momentum to them.

mv₁ + MV = mv₀

V = (mv₀ - mv₁)/M = (v₀-v₁)m/M

In other words, the final speed of the block is the speed the bullet lost,

but reduced by the ratio of masses.

Substituting actual numbers

V = (260 - 215)×23/5000 = 0.2 m/s