Question

The momentum of a body in two perpendicular direction at any time $t$ are given by $P_x=2t^2+6$ and $P_y=\frac{3t^2}{2}+3$. The force acting on the body at $t=2sec$ is given by

• A. $2units$
• B. $10units$
• C. $7units$
• D. $5units$

Answer 1

The correct option is B

$10units$

Step 1: Given Data

Momentum of body along x-axis: $P_x=2t^2+6$

Momentum of body along y-axis: $P_y=\frac{3t^2}{2}+3$

Time: $t=2sec$

Step 2: Formula Used

Force $\left(F\right)$ is the rate of change of momentum $\left(p\right)$ and is mathematically given as $F=\frac{dp}{dt}$.

Resultant force is calculated as $F=\sqrt{{F_x}^2+{F_y}^2}$, where $F_x$ is force along x-axis and $F_y$ is force along y-axis.

Step 3: Calculate the Resultant Force

Calculate the force along x-axis.

$\begin{array}{rcl}{F}_x& =& \frac{d{P}_x}{dt}\\ & =& \frac{d}{dt}\left(2t^2+6\right)\\ & =& 4t\\ & =& 4\times 2\\ & =& 8units\end{array}$

Calculate the force along y-axis.

$\begin{array}{rcl}{F}_y& =& \frac{d{P}_y}{dt}\\ & =& \frac{d}{dt}\left(\frac{3}{2}{t}^2+3\right)\\ & =& 3t\\ & =& 3\times 2\\ & =& 6units\end{array}$

Calculate the resultant force.

$\begin{array}{rcl}F& =& \sqrt{{F_x}^2+{F_y}^2}\\ & =& \sqrt{8^2+6^2}\\ & =& \sqrt{64+36}\\ & =& \sqrt{100}\\ & =& 10units\end{array}$

Hence, option B is correct.