Question

A particle of mass $m$ is projected with a velocity $6\hat{i}+8\hat{j}$. Then the magnitude of change in momentum when it just touches the ground will be

• A. $0$
• B. $12m$
• C. $16m$
• D. $20m$

Answer 1

The correct option is C

$16m$

Step. 1 Given Data,

A particle of mass $m$ projected with a velocity, $v$= $6\hat{i}+8\hat{j}$

Step. 2 Formula used,

$\Delta p=m(v_f-v_i)$

$\Delta p$is the change in momentum

$m$ is the mass of a particle.

$v_i$ is an initial velocity

$v_f$ is a final velocity

Step. 3 Calculation of magnitude of change in momentum,

Let us consider a particle of mass $m$ is projected with an initial velocity, $v_i$​ = $6\hat{i}+8\hat{j}$​

Since the particle is coming back to its initial position (i.e. to the level ground) we can say that its final velocity is also $6\hat{i}+8\hat{j}$. But the final velocity will be in a direction opposite to that of the initial vertical velocity.

Let us consider a particle of mass $m$ is projected with a final velocity, $v_f=6\hat{i}-8\hat{j}$

Change in momentum,

$$\begin{gathered} \Delta p=m(v_f-v_i)=m(6\hat{i}-8\hat{j}-(6\hat{i}+8\hat{j}\left)\right) \\ \Delta p=-16m\hat{j} \\ \left|\Delta p\right|=16m \end{gathered}$$

Therefore the magnitude of change in momentum is$16m$ when it just touches the ground.
Hence the correct option is C