Question

A bullet with a mass ????b=13.5 g is fired into a block of wood at velocity ????b=249. 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

Answer:

$M = 0.436 kg$

Explanation:

As per energy conservation we can say that energy stored in the spring at the position of maximum compression must be equal to the kinetic energy of bullet and block system

so here we have

$\frac{1}{2}(m + M)v^2 = \frac{1}{2} kx^2$

here we know that

k = 205 N/m

x = 35 cm

$\frac{1}{2}(m + M)v^2 = \frac{1}{2}(205)(0.35)^2$

now by momentum conservation we know that

$mv_o = (m + M)v$

$v = \frac{m}{m + M} v_o$

now plug in all values in it

$v = \frac{0.0135}{0.0135 + M}(249)$

now from above equation

$\frac{1}{2}(0.0135 + M)( \frac{0.0135}{0.0135 + M}(249))^2 = 12.56$

$\frac{5.65}{0.0135 + M} = 12.56$

by solving above equation we have

$M = 0.436 kg$