A particle of mass 4m explodes into three pie | vedclass

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(C) Let the third mass particle $(2m)$ move with velocity $u$ at an angle $	heta$ with the $X$-axis.The horizontal component of the velocity of the $

Vedclass · Accepted Answer

$\sqrt{13} 	ext{ ms}^{-1}$

Vedclass · Answer

(C) Let the third mass particle $(2m)$ move with velocity $u$ at an angle $	heta$ with the $X$-axis.The horizontal component of the velocity of the $2m$ mass particle is $u_x = u \cos 	heta$ and the vertical component is $u_y = u \sin 	heta$.Since the initial particle was at rest,the total linear momentum of the system must be zero.Applying the law of conservation of linear momentum in the $X$-direction:$0 = m(4) + 2m(u \cos 	heta)$$4m = -2m(u \cos 	heta)$$u \cos 	heta = -2 \quad \dots (i)$Applying the law of conservation of linear momentum in the $Y$-direction:$0 = m(6) + 2m(u \sin 	heta)$$6m = -2m(u \sin 	heta)$$u \sin 	heta = -3 \quad \dots (ii)$Squaring equations $(i)$ and $(ii)$ and adding them:$(u \cos 	heta)^2 + (u \sin 	heta)^2 = (-2)^2 + (-3)^2$$u^2(\cos^2 	heta + \sin^2 	heta) = 4 + 9$$u^2 = 13$$u = \sqrt{13} 	ext{ ms}^{-1}$

$A$ particle of mass $4m$ explodes into three pieces of masses $m, m$ and $2m$. The equal masses move along the $X$-axis and $Y$-axis with velocities $4 \text{ ms}^{-1}$ and $6 \text{ ms}^{-1}$ respectively. The magnitude of the velocity of the heavier mass is

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