Question

A body of mass $m=5\text{kg}$ is thrown at an angle $\theta ={30}^{\circ }$ to the horizontal with the initial velocity $v_0=20\text{m/s}$. Assuming the air drag to be negligible, find the magnitude of the momentum increment $\Delta p$ during the total time of motion.

• A. $0\text{kg m/s}$
• B. $50\text{kg m/s}$
• C. $100\text{kg m/s}$
• D. $10\text{kg m/s}$

Answer 1

The correct option is C $100\text{kg m/s}$

As we know that,
${\overrightarrow{F}}_{\text{ext}}=\frac{d\overrightarrow{p}}{dt}$
Here, ${\overrightarrow{F}}_{\text{ext}}=-\left(mg\right)\hat{j}=$ constant
$\Rightarrow F=\frac{dp}{dt}=\frac{\Delta p}{\Delta t}$ (if $F$ is constant) at $t=0$
Here, $mg=\frac{\Delta p}{\Delta t}\Rightarrow \Delta p=mg\left(\Delta t\right)$ where, $\Delta t$ is the Time of flight
or $\Delta p=mg\frac{2V_0\sin \theta }{g}$
Or, $\Delta p=2m{v}_0\sin \theta =2m{v}_0\sin {30}^{\circ }$
Or, $\Delta p=m{v}_0=5\times 20=100\text{kg m/s}$