A block moving horizontally on a smooth surface with a speed of 40 ms ⁻¹ splits into two equal parts. If one of the parts moves at 60 ms ⁻¹ in the same direction, then the fractional change in the kinetic energy will be x: 4 where x = ___________. Numerical answer type. Work it out, then open the answer.

Answer: 1. Equal parts m each: 2m(40) = m(60) + m v ⇒ v = 20 m/s Ki = 1/2(2m)(40)² = 1600m, Kf = 1/2m(60² + 20²) = 2000m (Δ K)/(Ki) = (400)/(1600) = 1/4 = x: 4 x = 1

A block moving horizontally on a smooth surface with a speed of 40 ms… | JEE Main 2021 (31 Aug, Shift 1)

Q7JEE Main 2021 · 31 AUG · SHIFT 1Medium

Collisions

A block moving horizontally on a smooth surface with a speed of 40 ms ${}^{-1}$ splits into two equal parts. If one of the parts moves at 60 ms ${}^{-1}$ in the same direction, then the fractional change in the kinetic energy will be x: 4 where x = ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 1

Equal parts $m$ each: $2m(40) = m(60) + m\,v \Rightarrow v = 20\text{ m/s}$
$K_i = \frac{1}{2}(2m)(40)^2 = 1600m$ , $K_f = \frac{1}{2}m(60^2 + 20^2) = 2000m$
$\frac{\Delta K}{K_i} = \frac{400}{1600} = \frac{1}{4} = x : 4$
$x = 1$

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