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Q.

A body of mass 10 kg is projected at an angle of 45° with the horizontal. The trajectory of the body is observed to pass through a point (20, 10). If T is the time of flight, then its momentum vector, at time $t = \frac{T}{\sqrt{2}}$ , is __________ [Take g = 10 m/s²]

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a

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

b

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

c

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

d

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

answer is D.

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Detailed Solution

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$\begin{array}{l} y = x - \frac{10 x^{2}}{2 u^{2} (\frac{1}{2})} \Rightarrow 10 = 20 - \frac{(10) (100)}{u^{2}} \\ u = 20 \\ T = \frac{(2) (20)}{\sqrt{2} (10)} = 2 \sqrt{2} \\ at t = 2 \sec \vec{v} = 10 \sqrt{2} \hat{i} + (10 \sqrt{2} - 10 (2)] \hat{j} \end{array}$

Momentum $\vec{p} = mv = 100 \sqrt{2} \hat{i} + (100 \sqrt{2} - 200) \hat{j}$

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