A mass of 2.9 kg is suspended from a string | vedclass

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(B) Let $M = 2.9 \ kg$ and $m = 0.1 \ kg$. The combined mass is $M' = M + m = 3.0 \ kg$.Using the law of conservation of momentum for the collision:$m

Vedclass · Accepted Answer

$135$

Vedclass · Answer

(B) Let $M = 2.9 \ kg$ and $m = 0.1 \ kg$. The combined mass is $M' = M + m = 3.0 \ kg$.Using the law of conservation of momentum for the collision:$m v = (M + m) V$$0.1 	imes 150 = 3.0 	imes V$$V = 5 \ m/s$ (velocity of the combined mass just after collision).Now,using the law of conservation of energy between the lowest point and the point where the string makes an angle $	heta = 60^{\circ}$:$\frac{1}{2} M' V^2 = \frac{1}{2} M' v_{	heta}^2 + M' g l(1 - \cos 	heta)$$v_{	heta}^2 = V^2 - 2gl(1 - \cos 60^{\circ})$$v_{	heta}^2 = 5^2 - 2 	imes 10 	imes 0.5 	imes (1 - 0.5) = 25 - 5 = 20 \ m^2/s^2$.The tension $T$ in the string at angle $	heta$ is given by:$T - M' g \cos 	heta = \frac{M' v_{	heta}^2}{l}$$T = M' g \cos 	heta + \frac{M' v_{	heta}^2}{l}$$T = 3.0 	imes 10 	imes \cos 60^{\circ} + \frac{3.0 	imes 20}{0.5}$$T = 30 	imes 0.5 + 120 = 15 + 120 = 135 \ N$.

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