Question

A 7.00 g bullet, when fired from a gun into a 1.10 kg block of wood held in a vise, penetrates the block to a depth of 9.60 cm. This block of wood is next placed on a frictionless horizontal surface, and a second 7.00 g bullet is fired from the gun into the block. To what depth will the bullet penetrate the block in this case

Answer 1

Explanation:

The given data is as follows.

    Mass of bullet 1, $m_{b}$ = 7 g = $7 \times 10^{-3} kg$

   Mass of block of wood, M = 1.10 kg

   Bullet penetrates to depth, $x_{1}$ = 9.60 cm = $9.60 \times 10^{-2} m$

  Mass of bullet 2, $m_{B}$ = 7 g = $7 \times 10^{-3} kg$

  Velocity of bullet = v

Let us assume that depth of penetration of the bullet is $x_{2}$.

According to the law of conservation of energy,

            Kinetic energy = Potential energy

        Kinetic energy = $\frac{1}{2}mv^{2}$

and,       Potential energy = mgh

so, for bullet 1

           $\frac{1}{2}m_{b}v^{2}_{1} = m_{b}gx_{1}$ .......... (1)

For bullet 2,

     Kinetic energy of bullet 1 - kinetic energy of bullet 2 = Potential energy

 $\frac{1}{2}m_{b}v^{2}_{1} - \frac{1}{2}(M + m_{b} + m_{B})v^{2}_{2} = m_{b}gx_{2}$

Using equation (1), we get

      $m_{b}gx_{1} - \frac{1}{2}(M + m_{b} + m_{B})v^{2}_{2} = m_{b}gx_{2}$

        $x_{2} = \frac{x_{1}(M + m_{b})}{(M + m_{b} + m_{B})}$

Putting the given values into the above formula as follows.

        $x_{2} = \frac{x_{1}(M + m_{b})}{(M + m_{b} + m_{B})}$

                   = $\frac{9.60 \times 10^{-2}(1.10 + 7 \times 10^{-3})}{(1.10 + 7 \times 10^{-3} + 7 \times 10^{-3})}$

                   = 0.0953 m

or,                = 9.53 cm    (as 1 m = 100 cm)

Thus, we can conclude that the bullet will penetrate at a depth of 9.53 cm.