QJEEM25Phc0c2: JEE Main 2025 Physics - Class 11 JEEM25Phc0c2 Solution: A body of mass 4 kg is placed on a plane at a point having…

$\begin{aligned}& \langle\mathrm{p}\rangle=\frac{(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \cdot(3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}})}{4}=6 \\& \overrightarrow{\mathrm{a}}=\left(\frac{\overrightarrow{\mathrm{F}}}{\mathrm{~m}}=\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \\& \overrightarrow{\mathrm{v}} \text { at } \mathrm{t}=4 \mathrm{sec}=\left(\frac{1}{2} \hat{\mathrm{i}}+\frac{3}{4} \hat{\mathrm{j}}\right) \times 4=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \\& P_{\text {ins }}=(2 \hat{\mathrm{i}}+3)(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})=13 \\& \frac{\langle\mathrm{P}\rangle}{P_{\text {ins }}}=\frac{6}{13}\end{aligned}$

Note: Given data is not matching.

$\begin{aligned}& S=u t+\frac{1}{2} a t^2 \\& S=0+\frac{1}{2} \frac{(2 \hat{i}+3 \hat{j})}{4}(4)^2=4 \hat{i}+6 \hat{\mathrm{j}}\end{aligned}$

If $\overrightarrow{\mathrm{r}}_{\mathrm{i}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}$ then $\overrightarrow{\mathrm{r}}_{\mathrm{f}}=7 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}$

But Final position given in the question is $(6,10)$ .