Respuesta :
Answer:
[tex]x=0.478\ m[/tex] is the compression in the spring
Explanation:
Given:
- mass of the bullet, [tex]m=10\ g=0.01\ kg[/tex]
- mass of block, [tex]M=2\ kg[/tex]
- stiffness constant of the spring, [tex]k=19.6\ N.m^{-1}[/tex]
- initial velocity of the spring just before it hits the block, [tex]u=300\ m.s^{-1}[/tex]
Now since the bullet-mass gets embed into the block, we apply the conservation of momentum as:
[tex]m.u=(M+m).v[/tex]
[tex]0.01\times 300=(2+0.01)\times v[/tex]
[tex]v=1.4925\ m.s^{-1}[/tex]
Now this kinetic energy of the combined mass gets converted into potential energy of the spring.
[tex]\rm Kinetic\ energy=Spring\ potential\ energy[/tex]
[tex]\frac{1}{2} (M+m).v^2=\frac{1}{2} k.x^2[/tex]
[tex]\frac{1}{2} \times (2+0.01)\times 1.4925^2=\frac{1}{2} \times 19.6\times x^2[/tex]
[tex]x=0.478\ m[/tex] is the compression in the spring
