A massless spring of force constant `1000 Nm^(-1)` is compressed a distance of `20 cm` between discs of `8 kg` and `2 kg`, spring is not attached to discs. The system is given an initial velocity `3 ms^(-1)` perpendicular to length of spring as shown in the figure. What is ground frame velocity of `2 kg` block (in `ms^(-1)`) when spring regains its natural length.

Answer

5

Explanation

Let `v_(1)` and `v_(2)` are the velocities along the spring of `A` and `B` when spring regains it natural length then
`2v_(1)=8v_(2)`…….i
`1/2 2v_(1)^(2)+1/28v_(2)^(2)=1/21000(20/100)^(2)`…………ii
Form i and ii `v_(1)=4ms^(-1), v_(2)=1ms^(-1)`
Ground frame velocity of `A`
`sqrt(v^(2)+v_(1)^(2))=sqrt(3^(2)+4^(2))=5ms^(-1)`