Answer:
Distance the spring is compressed, x = 0.52 m
Explanation:
Given :
Mass of the bullet, m = 10 g = 0.01 kg
Mass of the block, M = 1.95 kg
spring force constant, k = 16.6 N/m
Distance the spring is compressed =x
Speed of the bullet, v = 300 m/s
Speed of the block = V
Therefore we know that according to the law of conservation of momentum,
m.v = ( m+M )V
or [tex]V = \frac{m\times v}{m+M}[/tex]
[tex]V = \frac{0.01\times 300}{0.01+1.95}[/tex]
= 1.53 m/s
Now according to the law of conservation of momentum,
[tex]\frac{1}{2}\times M\times V^{2} =\frac{1}{2}\times k\times x^{2}[/tex]
[tex]x = \sqrt{\frac{\left ( M+m \right ).V^{2}}{k}}[/tex]
[tex]x = \sqrt{\frac{\left ( M+m \right )}{k}}\times V[/tex]
[tex]x = \sqrt{\frac{\left ( 1.95+0.01 \right )}{16.6}}\times 1.53[/tex]
[tex]x = 0.52[/tex] m
Thus, distance the spring is compressed, x = 0.52 m
