Question

A man of mass $70\text{kg}$ starts moving on the earth and aquires a velocity of $1\text{m/s}$. The recoiling speed of the earth is
(Assume mass of the earth is $6\times {10}^{24}\text{kg}$ and both earth and man were at rest initially)

• A. $1.16\times {10}^{-23}\text{m/s}$
• B. $11.6\times {10}^{-23}\text{m/s}$
• C. $1.6\times {10}^{-25}$ m/s
• D. $1.16\times {10}^{-24}$ m/s

Answer 1

The correct option is A $1.16\times {10}^{-23}\text{m/s}$

Given:
Mass of man; $m_1=70\text{kg}$
Mass of earth; $m_2=6\times {10}^{24}\text{kg}$
Final velocity of man; $v_1=1\text{m/s}$
Initial velocity of man ($u_1$) and earth ($u_2$) are zero as both are at rest. Let $v_2$ be the final velocity of earth.
Considering (man+earth) as system. As there is no net external force acting on the system.
$\therefore$ Applying principle of conservation of linear momentum (PCLM),
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
$0=70\left(1\right)+6\times {10}^{24}{v}_2$
$v_2=-1.16\times {10}^{-23}$ $\text{m/s}$
Here negative (- ve) sign shows that earth moves in the opposite direction of man.
$\therefore \text{Recoil speed=1.16}\times {10}^{-23}\text{m/s}$