A bullet is fired into a block of wood with speed 250 m/s. The block is attached to a spring that has a spring of 200 N/m. The block embedded bullet compresses the spring a distance of 30.0 cm to the right. Determine the mass of the wooden block

Answer 1

Assuming that the average bullet is 0.01kg
Pi=Pf
MiVi=(m+M)Vf so (0.01)(250)=(0.01+M)Vf
Vf= 2.5/(0.01+M)

PE(spring)=KE(Block+Bullet)
1/2Kx^2=1/2(0.01+M)V^2
Now Replace V with the solution from above:
1/2(200)(.3)^2 = 1/2(0.01+M)[2.5/(0.01+M)]^2
9J=1/2(0.01+M)[2.5/(0.01+M)]^2

So now you need to narrow it down to just one M to solve. So distribute the "^2" to the top and bottom terms.

9J=1/2(0.01+M)[(2.5^2)/(0.01+M)^2]
9J=0.5 * 6.25/(0.01+M)
0.01+M = 0.5 * 6.25/9
M= .337Kg is the block of wood