The block of mass 1 kg is at x = 10 m and moving along negative x-axis with velocity 6 `ms^(-1)`. Another block of mass 2 kg is at x = 12 m and moving along +ve x-axis with velocity `4ms^(-1)` at the same instant. The position of their centre of mass after 2s is

Options

A. 13 m
B. 12.67 m
C. 11.12 m
D. 14.77 m

Answer

B

Explanation

`x_(COM)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))=(1(10)+2(12))/(1+2)=(34)/3m`
`bar(v)_(COM)=m_(1)bar(v)_(1)+m_(2)bar(v)_(2))/(m_(1)+m_(2))=(1(-6)+2(4))/(1+2)=2/3ms^(-1)`
`bar(S)=bar(v)_(COM) times t=2/3 times 2=4/3m" (right wards)"`
`therefore X_(fCOM)=X_(iCOM)+bar(S)=34/3+4/3=38/3m=12.67m`