Three particles A, B and C of mass m, 2m and 4m respectively are placed in a straight line as shown. The distance of the centre of mass of the system of the three
particles from A and B respectively is.
Answers
Answered by
1
Answer:
Case-1
Let com be M(x,y).
x=
m+2m+3m+4m
m×0+2m×0+3m×l+4ml
=
10m
7ml
=
10
7
l
Similarly,
y=
10m
m×0+2m×l+3m×l+4m×0
=
10m
5ml
=
10
5
l
Case-2ABCD→BCDA
Let new position of com is N(x,y)
x=
10m
2m×0+3m×0+4m×l+m×l
=
10m
5ml
=
10
5
l
y=
10m
2m×0+3m×l+4m×l+m×0
=
10m
7ml
=
10
7
l
Distance between M and N is
(
10
7
−
10
5
)
2
+(
10
5
−
10
7
)
2
l
2
=
⎝
⎛
(
10
2
)
2
+(
10
2
)
2
⎠
⎞
l
d=
10
2
2
l=
5
2
l
Similar questions