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A pendulum bob of mass m connected at the end of an ideal string of length l is released from rest from horizontal position as shown in figure. At the lowest point, the bob makes an head on elastic collision with a stationary block of mass 5m, which is kept on a frictionless surface. Mark the incorrect statement(s) for the instant just after the impact. Acceleration due to gravity is g

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Tension in the string is $\frac{17 m g}{9}$

Tension in the string is 3 mg

Magnitude of velocity of the block is $\frac{\sqrt{2 gl}}{3}$

The maximum height attained by pendulum bob after the impact (measured from lowest position) is $\frac{4 l}{9}$

answer is B.

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Detailed Solution

Speed of the bob at its lowest point is $\sqrt{2 gl}$

After impact velocity of bob,

$V_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) u_{1} = (\frac{m - 5 m}{m + 5 m}) \sqrt{2 gl} = \frac{- 2}{3} \sqrt{2 gl}$

After impact velocity of block,

$v_{2} = \frac{2 m_{1} u_{1}}{m_{1} + m_{2}} = \frac{2 m}{6 m} \sqrt{2 gl} = \frac{\sqrt{2 gl}}{3}$

*Height obtained by bob after collision
$H = \frac{V_{1}^{2}}{2 g} = \frac{\frac{4}{9} 2 gl}{2 g} = \frac{4 l}{9}$

*Tension in the string is $T - mg = \frac{{mV}_{1}^{2}}{l}$
$T = mg + \frac{m}{l} {(\frac{2}{3} \sqrt{2 gl})}^{2} = mg + \frac{8 mg}{9} = \frac{17 mg}{9}$

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