Explanation:
As the given collision is inelastic in nature. Therefore, momentum will be conserved. Also, both the block and bullet have the same final velocity.
Therefore, formula will be as follows.
[tex]m_{1}v_{1} + m_{2}v_{2} = V_{f}(m_{1} + m_{2})[/tex]
Also,
[tex]\frac{1}{2}kx^{2} = \frac{1}{2}V^{2}_{f}(m_{1} + m_{2})[/tex]
where, k = spring constant
x = compression of spring
[tex]m_{1}[/tex] = mass of bullet
[tex]m_{2}[/tex] = mass of wooden block
Putting the given values into the above formula as follows.
[tex]\frac{1}{2}kx^{2} = \frac{1}{2}V_{f}^{2}(m_{1} + m_{2})[/tex]
[tex]\frac{1}{2}150 N/m \times (0.8 m)^{2} = \frac{1}{2}V^{2}_{f}(12 g + 100 g)[/tex]
[tex]V_{f}[/tex] = 273.2 m/s
= 273 m/s (approx)
Therefore, we can conclude that speed of the bullet is coming with 273 m/s at impact with the block.
