Question

A 12.0-g bullet is fired horizontally into a 114-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 146 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 75.5 cm, what was the speed of the bullet at impact with the block

Answer 1

Answer:

The speed of the bullet-block is $V=0.81 ms^{-1}$.

Explanation:

The expression of the conservation of energy to the given system is as follows;

$KE_i+PE_i=KE_f+PE_f$

Here, $KE_i,KE_f$ are the initial kinetic energy and final kinetic energy and $PE_i,PE_f$.

$\frac{1}{2}mv^{2}+\frac{1}{2}kx^{2}=\frac{1}{2}(m+M)V^2+0$

Here, m is the mass of the bullet, M is the mass of the wooden block, k is the spring constant, x is the distance, v is the speed of the bullet and V is the speed of the bullet with the block.

Calculate the speed of the bullet-block.

Convert the distance from cm to m.

x=75.5 cm

x=0.755 m

Put m= 12 g, M=114 g, v=0, x= 0.755 m and $k=146 Nm^{-1}$ in the above expression according to the given values in the problem.

$\frac{1}{2}(12)(0)^{2}+\frac{1}{2}(146)(0.755)^{2}=\frac{1}{2}(12+114)V^2+0$

$\frac{1}{2}(146)(0.755)^{2}=\frac{1}{2}(12+114)V^2$

$V=0.81 ms^{-1}$

Therefore, the speed of the bullet-block is $V=0.81 ms^{-1}$.