Question

A 13.1-g bullet is fired into a block of wood at 261 m/s. The block is attached to a spring that has a spring constant of 205 N/m. The block and bullet continue to move, compressing the spring by 35.0 cm before the whole system momentarily comes to a stop. Assuming that the surface on which the block is resting is frictionless, determine the mass of the wooden block.

Answer 1

please help this is the only problem i do not know how to do!!!

Answer 2

well, the spring will lead you to the initial KE of the block
PE spring comptessed=initiaL KE of block/bullet combo.

from that
1/2 k x^2= (1/2 (M+m)v'^2)
M is block mass, m is bullet mass. Find v'
Now law of conservation of momentum at the collision
(M+m)v'=m*261
solve for M

Answer 3

Pi=Pf
MiVi=(m+M)Vf so (0.0131)(261)=( 0.0131 +M)Vf
Vf= 3.4191/( 0.0131 +M)

PE(spring)=KE(Block+Bullet)
1/2Kx^2=1/2(0.0131 +M)V^2
Now Replace V with the solution from above:
1/2(205)(.35)^2 = 1/2(0.0131 +M)[3.4191/( 0.0131 +M)]^2
12.56J= 1/2(0.0131 +M)[3.4191/( 0.0131 +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.

12.56J=1/2(0.0131 +M)[(3.4191^2)/( 0.0131 +M)^2]
12.56J=0.5 * 6.25/( 0.0131 +M)
0.0131 +M = 0.5 * 11.69/12.56
M= .452Kg is the block of wood

Answer 4

Disregard the "6.25" above it should also just be 11.69