Question 95: Centre of Mass and Rotational Motion

Question

A body of mass $$10 \mathrm{~kg}$$ is projected at an angle of $$45^{\circ}$$ with the horizontal. The trajectory of the body is observed to pass through a point $$(20,10)$$. If $$\mathrm{T}$$ is the time of flight, then its momentum vector, at time $$\mathrm{t}=\frac{\mathrm{T}}{\sqrt{2}}$$, is _____________.
[Take $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$$ ]

Accepted Answer

m = 10 kg
$$	heta$$ = 45$$^\circ$$
$$y = x	an 	heta \left( {1 - {x \over R}} ight)$$
$$ \Rightarrow 10 = 20\left( {1 - {{20} \over R}} ight)$$
$$ \Rightarrow R = 40$$
$$40 = {{{u^2}} \over {10}} \Rightarrow u = 20$$
$$ \Rightarrow T = {{20 	imes 20 	imes {1 \over {\sqrt 2 }}} \over {10}} = {4 \over {\sqrt 2 }}s \Rightarrow t = 2\,s$$
at $$t = 2,\,\overrightarrow v  = \left( {10\sqrt 2 \widehat i} ight) + \left( {10\sqrt 2  - 2 	imes 10} ight)\widehat j$$
$$ \Rightarrow \overrightarrow p  = 10\left[ {10\sqrt 2 \widehat i + \left( {10\sqrt 2  - 20} ight)\widehat j} ight]$$
$$ = 100\sqrt 2 \widehat i + \left( {100\sqrt 2  - 200} ight)\widehat j$$

Answer

$$
100 \hat{i}+(100 \sqrt{2}-200) \hat{j}$$

Answer

$$100 \sqrt{2} \hat{i}+(100-200 \sqrt{2}) \hat{j}$$

Answer

$$100 \hat{i}+(100-200 \sqrt{2}) \hat{j}$$

Answer

$$100 \sqrt{2} \hat{i}+(100 \sqrt{2}-200) \hat{j}$$

Centre of Mass and Rotational Motion

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