JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A body of mass \(10 kg\) 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 \(T\) is the time of flight, then its momentum vector, at time \(t =\frac{ T }{\sqrt{2}}\), is. \(\left[\right.\) Take \(\left.g=10 m / s ^{2}\right]\)
- A \(100 \hat{ i }+(100 \sqrt{2}-200) \hat{ j }\)
- B \(100 \sqrt{2} \hat{i}+(100-200 \sqrt{2}) \hat{j}\)
- C \(100 \hat{ i }+(100-200 \sqrt{2}) \hat{ j }\)
- D \(100 \sqrt{2} \hat{i}+(100 \sqrt{2}-200) \hat{j}\)
Answer & Solution
Correct Answer
(D) \(100 \sqrt{2} \hat{i}+(100 \sqrt{2}-200) \hat{j}\)
Step-by-step Solution
Detailed explanation
\(y = x -\frac{10 x ^{2}}{2 u ^{2}\left(\frac{1}{2}\right)} \Rightarrow 10=20-\frac{(10)(100)}{ u ^{2}}\) \(u=20\) \(T=\frac{(2)(20)}{\sqrt{2}(10)}=2 \sqrt{2}\) \(\overrightarrow{ v }=10 \sqrt{2} \hat{ i }+(10 \sqrt{2}-10(2)] \hat{ j }\) Momentum…
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