Two particles are shown in figure. At t=0 a constant force F=6 N starts acting on 3 kg. Find the velocity of centre of
mass of these particles at t= 5 s.
>
2 kg
3 kg
F=9N
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Explanation:
We know that the total linear momentum of a system of partilas is equal to linar momentumm of center of man of the system.
⇒(m
1
+m
2
)ϑ
cm
=m
1
ϑ
1
+m
2
ϑ
2
At t= 5 sec,
ϑ
1
=0m/s
ϑ
2
=u
2
+at
ϑ
2
=2×5=10m/s(∵u
2
=0 and a=
m
2
F
=2m/s
2
)
⇒5×ϑ
cm
=0+3×10
⇒ϑ
cm
=6m/s
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