TS EAMCET · Physics · Center of Mass Momentum and Collision
A bouncing ball of mass \(200 \mathrm{~g}\) falls from the height of \(5 \mathrm{~m}\) on a horizontal ground. After every impact with the ground, the velocity of the ball decreases by \(\frac{1}{2}\) times. The total momentum, the ball imparts on to the ground after 3 impacts is \(\left(\right.\) Let \(\left.g=10 \mathrm{~m} / \mathrm{s}^2\right)\)
- A \(\frac{14}{4} \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)
- B \(\frac{20}{6} \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)
- C \(\frac{26}{12} \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)
- D \(\frac{21}{4} \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(D) \(\frac{21}{4} \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Given, mass of ball, \(m=200 \mathrm{~g}=0.2 \mathrm{~kg}\) and height of the ball, \(h=5 \mathrm{~m}\) \(\therefore\) Velocity of ball at the time of first impact, \(v_1^{\prime}=\frac{v_1}{2}=\frac{10}{2}=5 \mathrm{~m} / \mathrm{s}\) \(\therefore\) Momentum imparted by the…
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