Two objects having masses $1: 4$ ratio are at rest. When both of them are subjected to same force separately…
- $4$
- $2$
- $2.5$
- $1$
Solution
After time $t_2$, $\begin{aligned} & \mathrm{k}_2=\frac{1}{2}(4 \mathrm{~m}) \mathrm{v}_2^2=\frac{1}{2}(4 \mathrm{~m})\left(\mathrm{at}_2\right)^2=2 \mathrm{ma}^2 \mathrm{t}_2^2 \\ & \therefore \mathrm{~K}_1=\mathrm{K}_2 \Rightarrow \frac{1}{2} \mathrm{ma}^2 \mathrm{t}_1^2=2 \mathrm{ma}^2 \mathrm{t}_2^2 \quad \therefore \frac{\mathrm{t}_1}{\mathrm{t}_2}=2 \end{aligned}$
Asked in: AP EAMCET 2024 (20 May Shift 2)