Answer: 0.43 m
Explanation:
Given
mass of bullet [tex]m=10\ gm\approx 0.01\ kg[/tex]
mass of block [tex]M=1.95\ kg[/tex]
The Force constant of spring is [tex]k=23.9\ N/m[/tex]
Speed of bullet is [tex]u=300\ m/s[/tex]
Conserving the energy i.e. kinetic energy of the bullet and box is converted into Elastic potential energy of spring
[tex]\Rightarrow \dfrac{1}{2}(M+m)v^2=\dfrac{1}{2}kx^2[/tex]
Conserving linear momentum
[tex]\Rightarrow mu=(M+m)v\\\\\Rightarrow v=\dfrac{mu}{M+m}[/tex]
Put the value of [tex]v[/tex] we get
[tex]\Rightarrow x=mu\sqrt{\dfrac{1}{k(M+m)}}\\\Rightarrow x=0.01\times 300\sqrt{\dfrac{1}{23.9(0.01+1.95)}}\\\Rightarrow x=3\sqrt{\dfrac{1}{46.844}}=\dfrac{3}{6.844}=0.43\ m[/tex]
Thus, spring will be compressed to a distance of [tex]0.43\ m[/tex]
