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 initial velocity of the combined mass $V$ after collision is found using the conservation of linear mom

Vedclass · Accepted Answer

$135$

Vedclass · Answer

(B) Let $M = 2.9 \ kg$ and $m = 0.1 \ kg$. The initial velocity of the combined mass $V$ after collision is found using the conservation of linear momentum:$M(0) + m(150) = (M + m)V$$0.1 	imes 150 = (2.9 + 0.1)V$$15 = 3V \Rightarrow V = 5 \ m/s$.Now,let the velocity of the combined mass be $v$ at an angle $	heta = 60^{\circ}$. By conservation of mechanical energy:$\frac{1}{2}(M+m)V^2 = \frac{1}{2}(M+m)v^2 + (M+m)gL(1 - \cos 	heta)$$\frac{1}{2}(5)^2 = \frac{1}{2}v^2 + 10 	imes 0.5 	imes (1 - \cos 60^{\circ})$$12.5 = 0.5v^2 + 5 	imes (1 - 0.5) = 0.5v^2 + 2.5$$0.5v^2 = 10 \Rightarrow v^2 = 20 \ m^2/s^2$.The tension $T$ in the string at angle $	heta$ is given by:$T - (M+m)g \cos 	heta = \frac{(M+m)v^2}{L}$$T = (M+m) \left( g \cos 	heta + \frac{v^2}{L} ight)$$T = 3 \left( 10 	imes \cos 60^{\circ} + \frac{20}{0.5} ight)$$T = 3 \left( 10 	imes 0.5 + 40 ight) = 3 	imes 45 = 135 \ N$.

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