Question: In a system of three particles, the linear momenta of the three particles are (1,2,3), (4,5,6) and(5,6,7). These components are in kgms–1. If the velocity of centre of mass of the system is(30,39,48) ms–1, then find the total mass of the system.
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Answer:
We can analyze the situation as mass 6kg is at (1,2,3), another mass of 5kg at (−1,−2,0). We have to find the position of other mass of 5kg to be put say (x,y,z) so that the final center of mass is at (1,2,3)
x
CM
=
m
1
+m
2
+m
3
m
1
x
1
+m
2
x
2
+m
3
x
3
⟹1=
16
6×1+5×(−1)+5×x
⟹x=3m
y
CM
=
m
1
+m
2
+m
3
m
1
y
1
+m
2
y
2
+m
3
y
3
⟹2=
16
6×2+5×(−2)+5×y
⟹y=6m
z
CM
=
m
1
+m
2
+m
3
m
1
z
1
+m
2
z
2
+m
3
z
3
⟹3=
16
6×3+5×0+5×z
⟹z=6m
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