In Fig. 9-59, a 10 gbullet moving directly upward at 1000 m/sstrikes and passes through the center of mass of a 5.0 kgblock initially at rest. The bullet emerges from the block moving directly upward at 400 m/s. To what maximum height does the block then rise above its initial position?

Mass of the bullet is, ${\mathrm{m}}_{\mathrm{bu}}=10\mathrm{g}=0.01\mathrm{kg}$

Mass of the block is,${\mathrm{m}}_{bl}=5\mathrm{kg}$

Final speed of the bullet is, $v_{fb}=400\mathrm{m}/\mathrm{s}$

Initial speed of the block is, $v_{ib}=0\mathrm{m}/\mathrm{s}$

By applyingtheprinciple of conservation of momentum and usingtheconcept of conversion of kinetic energy to potential energy at maximum height, findtherise in the height of the block. According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formulae are as follow:

$$\begin{gathered} P_i=P_f \\ P=mv \end{gathered}$$

At maximum height,$\frac{1}{2}m{v}^2=mgh$

Where, m is mass, v is velocity, h is height, g is an acceleration due to gravity, P is linear momentum.

To calculate the rise in the height of the block (h), use theprinciple of conservation of momentum.

Total momentum$\overrightarrow{P_i}$before collision = Total momentum after collision$\overrightarrow{P_f}$

For the given situation,

Total initial momentum = Initial momentum of bullet + Initial momentum of block

$\overrightarrow{P_1}=\overrightarrow{P_{i\left(bu\right)}}+\overrightarrow{P_{1\left(bl\right)}}$

As initially block is at rest,$\overrightarrow{P_{1\left(bl\right)}}=0$

Total final momentum = final momentum of bullet + final momentum of block

$\overrightarrow{P_f}=m_{bu}{v}_{f\left(bu\right)}+m_b{v}_{f\left(bl\right)}.....\left(2\right)$

Equating equation (1) and (2),

$m_{bu}{v}_{i\left(bu\right)}+m_{bu}{v}_{f\left(bu\right)}+m_{bl}{v}_{f\left(bl\right)}$

Final velocity of the block can be calculated by,

$$\begin{gathered} v_{f\left(bl\right)}=\frac{m_{bu}{v}_{i\left(bu\right)}-m_{bu}{v}_{f\left(bu\right)}}{m_{bl}} \\ {v}_{f\left(bl\right)}=\frac{0.01\times 1000-0.01\times 400}{5} \end{gathered}$$

The final velocity of blockis 1.2 m/s.

To find the rise in the height of the block, use $v_{f\left(bl\right)}$

At maximum height, $\frac{1}{2}m{v}^2=mgh$

Cancelling mass m and rearranging the equation for h,

$h=\frac{{\left(v_{f\left(bl\right)}\right)}^2}{2\left(g\right)}$

Substituting the values in the above equation,

$$\begin{gathered} h=\frac{{\left(1.2\right)}^2}{2(-9.8)} \\ h=0.073\mathrm{m} \end{gathered}$$

Hence, rise in the height of the block is, h = 0.073 m.

Therefore, by applyingtheprinciple of conservation of momentum and usingtheconcept of conversion of K.E. to P.E. at maximum height, the height of the block (h) can be calculated.