TS EAMCET · Physics · Motion In Two Dimensions
A ball of mass \(0.2 \mathrm{~kg}\) is thrown from a height of \(1 \mathrm{~m}\) and with an initial velocity of \(\sqrt{10} \mathrm{~m} / \mathrm{s}\) at an angle of \(45^{\circ}\) with the horizontal. Assuming, acceleration due to gravity \(g=10 \mathrm{~m} /\) \(\mathrm{s}^2\), then modulus of momentum increment during the total time of motion in \(\mathrm{kg} \mathrm{m} / \mathrm{s}\) is
- A \(\frac{2+\sqrt{10}}{\sqrt{10}}\)
- B \(\frac{1+\sqrt{10}}{\sqrt{5}}\)
- C \(\frac{1+\sqrt{5}}{\sqrt{5}}\)
- D \(\frac{\sqrt{5}-1}{\sqrt{5}}\)
Answer & Solution
Correct Answer
(C) \(\frac{1+\sqrt{5}}{\sqrt{5}}\)
Step-by-step Solution
Detailed explanation
There is no change in velocity in \(x\)-direction, so momentum change occurs only in \(y\)-direction. Now initial momentum in \(y\) direction is \( \mathrm{p}_y(\text { initial })=m v_y=0.2\left(\sqrt{10} \sin 45^{\circ} \hat{\mathbf{j}}\right) \)…
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