JEE MAIN - Physics (2024 - 30th January Morning Shift - No. 8)
A particle is placed at the point $$A$$ of a frictionless track $$A B C$$ as shown in figure. It is gently pushed towards right. The speed of the particle when it reaches the point B is :
(Take $$g=10 \mathrm{~m} / \mathrm{s}^2$$).
_30th_January_Morning_Shift_en_8_1.png)
$$2 \sqrt{10} \mathrm{~m} / \mathrm{s}$$
$$10 \mathrm{~m} / \mathrm{s}$$
$$\sqrt{10} \mathrm{~m} / \mathrm{s}$$
$$20 \mathrm{~m} / \mathrm{s}$$
Explanation
By COME
$$\begin{aligned} & \mathrm{KE}_{\mathrm{A}}+\mathrm{U}_{\mathrm{A}}=\mathrm{KE}_{\mathrm{B}}+\mathrm{U}_{\mathrm{B}} \\ & 0+\mathrm{mg}(1)=\frac{1}{2} \mathrm{mv}^2+\mathrm{mg} \times 0.5 \\ & v=\sqrt{g}=\sqrt{10} \mathrm{~m} / \mathrm{s} \end{aligned}$$
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