Answer:
The speed of the bullet at impact equals 57.44 m/s
Explanation:
Since the surface is friction less thus no energy shall be lost in the system
Equating the initial kinetic energy of the bullet with the final potential energy of the compressed spring we get
[tex]E_{i}=\frac{1}{2}mv^{2}[/tex]
Potential Energy of the compressed spring is given as
[tex]E_{f}=\frac{1}{2}k(\delta x)^{2}[/tex]
given
k = 120 N/m
m = 11 grams
x = 55 cm
Equating the 2 energies and solving for velocity we get
[tex]\frac{1}{2}mv^{2}=\frac{1}{2}k(\delta x)^{2}\\\\v^{2}=\frac{k(\delta x)^{2}}{m}\\\\v^{2}=\frac{120\times (0.55)^{2}}{11\times 10^{-3}}\\\\v^{2}=3300\\\\\therefore v=57.44m/s[/tex]
