QJEEM19Ph6fc1: JEE Main 2019 Physics - Class 11 JEEM19Ph6fc1 Solution: A body of mass 1 kg falls freely from a height of 100 m, on a…

velocity of 1 kg block just before it collides with 3kg block

= $$\sqrt{2gh}=\sqrt{2000}$$ m/s

Applying momentum conversation just before and just after collision.

1 $$\times$$ $$\sqrt{2000}$$ = 4v $$\Rightarrow$$ v = $$\frac{\sqrt{2000}}{4}$$ m/s



initial compression of spring

1.25 $$\times$$ 10⁶ x₀ = 30 $$\Rightarrow$$ x₀ $$\approx$$ 0

applying work energy theorem,

W_g + W_sp = $$\mathrm{\Delta }$$KE

$$\Rightarrow$$   40 $$\times$$ x + $$\frac{1}{2}$$ $$\times$$ 1.25 $$\times$$ 10⁶ (0² $$-$$ x²)

= 0 $$-$$ $$\frac{1}{2}$$ $$\times$$ 4 $$\times$$ v²

solving x $$\approx$$ 4 cm