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

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Conservation of Momentum

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 =

_____.

Answer: 1

Solution

Momentum is conserved in the split:

m (40) = \frac{m}{2} (60) + \frac{m}{2} v_{2}

gives

v_{2} = 20 m s^{- 1}

. The initial kinetic energy is

K_{i} = \frac{1}{2} m (40)^{2} = 800 m

. The final kinetic energy is

K_{f} = \frac{1}{2} \frac{m}{2} (60)^{2} + \frac{1}{2} \frac{m}{2} (20)^{2} = \frac{m}{4} (3600 + 400) = 1000 m .

The fractional change is

\frac{K _{f} - K _{i}}{K _{i}} = \frac{200 m}{800 m} = \frac{1}{4}

. Written as

x : 4

this is

1 : 4

, so

x = 1

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Integer answer

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