JEE Main 11 Jan 2019 Shift 1 Solved Paper

Velocity of 1 kg block just before it collideswith 3kg block = √2gh = √2000 m/s
Applying momentum conversation just beforeand just after collision
1 × √2000 = 4v ⇒ v =

√2000

4

m/s

initial compression of spring
1.25 × 106x0 = 30 ⇒ x0 ~ 0
applying work energy theorem
Wg+Wsp = ΔKE
⇒ 40 × x +

1

2

× 1.55 × 106(02−x2)
= 0 -

1

2

×4×v2
solving x ~ 4 cm