91Ó°ÊÓ

Given that a 12.0-g rifle bullet is fired with a speed of 380 m/s into a ballistic pendulum with mass 6.00 kg, suspended from a cord 70.0 cm long.

Mass of bullet m = 12g

Initial speed v = 380 m/s

Mass of pendulum M = 6kg

Conservation of momentum states the if no external force is acting, then total momentum of the system is conserved.

(a)

After the bullet collides the pendulum, it is stuck in the pendulum.

Let their combined velocity is V and height to which it rises is h.

During collision, momentum is conserved.

So, momentum of bullet = momentum of combined bullet and pendulum

$$\begin{gathered} mv=(M+m)V \\ 12×{10}^{-3}kg×380m/s=(6kg+12×{10}^{-3}kg)V \\ V=0.758m/s \end{gathered}$$

So the velocity is 0.758 m/s.

Since the energy is also conserved, so

$$\begin{gathered} \frac{1}{2}M{V}^2=Mgh \\ h=\frac{V^2}{2g} \end{gathered}$$

So$h=\frac{{(0.758m/s)}^2}{2(9.8m/s^2)}=0.0293m$

Hence, it rises to a height of 2.93 cm

(b)

The initial kinetic energy of the bullet is

$$\begin{gathered} \frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}\left(12×{10}^{-3}\mathrm{kg}\right){\left(380\mathrm{m}/\mathrm{s}\right)}^2 \\ =866\mathrm{J} \end{gathered}$$

Hence the initial kinetic energy is 866J.

The kinetic energy of bullet and pendulum immediately after collision is

$$\begin{gathered} \frac{1}{2}(\mathrm{M}+\mathrm{m}){\mathrm{V}}^2=\frac{1}{2}\left(6\mathrm{kg}+12×{10}^{-3}\mathrm{kg}\right){\left(0.758\mathrm{m}/\mathrm{s}\right)}^2 \\ =1.73\mathrm{J} \end{gathered}$$

Hence the kinetic energy of bullet and pendulum immediately after collision is 1.73 J.