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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 threeparticles from A and B respectively is. |
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Answer» 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
) +( 10 5
− 10 7
) 2 l 2
= ⎝ ⎛
( 10 2
) 2 +( 10 2
) 2
⎠ ⎞
l d= 10 2 2
l= 5 2
l
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