Kinematics Ques 48
- A body is projected at $t=0$ with a velocity $10 ms^{-1}$ at an angle of $60^{\circ}$ with the horizontal. The radius of curvature of its trajectory at $t=1 s$ is $R$. Neglecting air resistance and taking acceleration due to gravity $g=10 ms^{-2}$, the value of $R$ is
(2019 Main, 11 Jan I)
(a) $10.3 m$
(b) $2.8 m$
(c) $5.1 m$
(d) $2.5 m$
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Answer:
Correct Answer: 48.(b)
Solution:
Formula:
- Components of velocity at an instant of time $t$ of a body projected at an angle $\theta$ is
$$ v _x=u \cos \theta+g _x t \text { and } v _y=u \sin \theta+g _y t $$
Here, components of velocity at $t=1 s$, is
$$ \begin{aligned} v _x & =u \cos 60^{\circ}+0 \quad\left[\text { as } g _x=0\right] \\ & =10 \times \frac{1}{2}=5 m / s \\ \text { and } v _y & =u \sin 60^{\circ}+(-10) \times(1) \\ & =10 \times \frac{\sqrt{3}}{2}+(-10) \times(1)=5 \sqrt{3}-10 \\ \Rightarrow \quad\left|v _y\right| & =|10-5 \sqrt{3}| m / s \end{aligned} $$
Now, angle made by the velocity vector at time of $t=1 s$
$$ \begin{array}{rlrl} & & |\tan \alpha| & =\left|\frac{v _y}{v _x}\right|=\frac{|10-5 \sqrt{3}|}{5} \\ \Rightarrow \quad & \tan \alpha=|2-\sqrt{3}| \text { or } \alpha=15^{\circ} \end{array} $$
$\therefore \quad$ Radius of curvature of the trajectory of the projected body
$$ \begin{aligned} & R=v^{2} / g \cos \alpha=\frac{(5)^{2}+(10-5 \sqrt{3})^{2}}{10 \times 0.97} \\ & {\left[\because v^{2}=v _x^{2}+v _y^{2} \text { and } \cos 15^{\circ}=0.97\right]} \\ & \Rightarrow \quad R=2.77 m \approx 2.8 m \end{aligned} $$
BETA
