Question

A stone of mass $7\text{kg}$ is dropped from a height of $5\text{m}$. Calculate the linear momentum of the stone when it reaches the ground.

Take $g=10{\text{m/s}}^2$

• A. $50\text{kg-m/s}$
• B. $70\text{kg-m/s}$
• C. $35\text{kg-m/s}$
• D. $25\text{kg-m/s}$

Answer 1

The correct option is B $70\text{kg-m/s}$

From law of conservation of mechanical energy, loss in potential energy is equal to gain in kinetic energy.

So,

$KE=mgh$

Now, linear momentum,

$p=\sqrt{2mKE}=\sqrt{2m\times mgh}=\sqrt{2m^{2}gh}$

$\Rightarrow p=\sqrt{2\times 7^2\times 10\times 5}=70\text{kg-m/s}$

​​​​​​​Hence, option $\left(B\right)$ is the correct answer.