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A mass of 3.9 kg suspended from a string of length 0.5m is at rest. Another body of mass 0.1kg moving horizontally with a velocity $200 {ms}^{- 1}$ strikes and sticks to it. The tension in the string when it makes $60 °$ with the vertical is $(g = 10 {ms}^{- 2})$

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180 N

220 N

80 N

100 N

answer is A.

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Detailed Solution

From L.C.L.M

$0.1 \times 200 = 4 \times v$

$v = 5 m/s$

$T - mgcos θ = \frac{{mv}_{0}^{2}}{r}$

$v_{0} = \sqrt{{(5)}^{2} - 2 g l (1 - \cos θ)}$

$= \sqrt{25 - 20 (\frac{1}{2}) (\frac{1}{2})}$

$= \sqrt{20}$

$T - 40 \times \frac{1}{2} = \frac{4 (20)}{1/2}$

$T - 20 = 8 (20)$

$T = 160 + 20$

$T = 180 N$

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