Centre of mass (C.M.) of three particles of masses 1 kg, 2 kg and 3 kg lies at the point (1,2,3) and C.M. of another system of particles of 3 kg and 2 kg lies at the point (-1,3,-2). Where should we put a particle of mass 5 kg so that the C.M. of entire system lies at the C.M. of the first system ?

Question

Centre of mass (C.M.) of three particles of masses 1 kg, 2 kg and 3 kg lies at the point (1,2,3) and C.M. of another system of particles of 3 kg and 2 kg lies at the point (-1,3,-2). Where should we put a particle of mass 5 kg so that the C.M. of entire system lies at the C.M. of the first system ? (A) (3,1,8) (B) (0,0,0) (C) (1,3,2) (D) (-1,2,3)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/39521fd2dfe287f71c0be90f47953fb2-.png" /> barx=(m1 x1+m2 x2/m1+m2) 1=(5(-1)+5 × x/10) -5+5 x=10 ⇒ x=3 bary=2=(5 × 3+5 y/10), 15+5 y=20, y=1 barz=3=(5 x-2+5 ⋅ z/10),-10+5 z=30, z=8

$(3, 1, 8)$

$(0, 0, 0)$

$(1, 3, 2)$

$(- 1, 2, 3)$

(3,1,8)

(0,0,0)

(1,3,2)

(−1,2,3)

xˉ=m1​+m2​m1​x1​+m2​x2​​ 1=105(−1)+5×x​ −5+5x=10⇒x=3 yˉ​=2=105×3+5y​,15+5y=20,y=1 zˉ=3=105x−2+5⋅z​,−10+5z=30,z=8

$(3, 1, 8)$

$(0, 0, 0)$

$(1, 3, 2)$

$(- 1, 2, 3)$