A 91 kgman lying on a surface of negligible friction shoves a 68 g stone away from himself, giving it a speed of4.0 m/s. What speed does the man acquire as a result?

1. The mass of the man,$m_1\mathrm{is}91\mathrm{kg}$ .
2. The mass of the stone, $m_2\mathrm{is}68\times {10}^{-3}\mathrm{kg}$.
3. The final speed of the stone, $v_2\mathrm{is}4.0\mathrm{m}/\mathrm{s}$.

The man and the stone are initially at rest. When the man shoves the stone, both acquire some velocity. This change in momentum of both, the man and stone, follows the law of conservation of linear momentum which is given as follows.

${\mathbf{m}}_{\mathbf{1}}{\mathbf{u}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{u}}_{\mathbf{2}}\mathbf{=}{\mathbf{m}}_{\mathbf{1}}{\mathbf{v}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{v}}_{\mathbf{2}}$

We apply the law of conservation of linear momentum to determine the final speed of the man.

$m_1{v}_{i1}+m_2{v}_{i2}=m_1{v}_{f1}+m_2{v}_{f2}$

Since the man and the stone are initially at rest,$v_{j1}=v_{i2}=0\mathrm{m}/\mathrm{s}$

$$\begin{gathered} 0=m_1{v}_{f1}+m_2{v}_{f2} \\ {m}_1{v}_{f1}=m_2{v}_{f2} \\ {v}_{f1}=\frac{-68\times {10}^{-3}\times 4.0}{91} \\ =3.0\times {10}^{-3}\mathrm{m}/\mathrm{s} \\ =-3.0\mathrm{mm}/\mathrm{s} \end{gathered}$$

The negative sign indicates that the man moves in the direction opposite to that of the stone. The magnitude of final speed of man is $v_{f1}=3.0\times {10}^{-3}\mathrm{m}/\mathrm{s}$