Question

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​

Answer 1

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Answer 2

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