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QJEEM21Ph62bc: JEE Main 2021 Physics - Class 11 JEEM21Ph62bc Solution: A block moving horizontally on a smooth surface with a speed of …

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 = ___________.

Step-by-Step Verified Solution

JEE Main 2021 (Online) 31st August Morning Shift Physics - Work Power & Energy Question 71 English Explanation
Pi = Pf

m ×\times 40 = m2{m \over 2} ×\times v + m2{m \over 2} ×\times 60

40 = v2{v \over 2} + 30

⇒\Rightarrow v = 20

(K. E.)I = 12{1 \over 2} m ×\times (40)2 = 800 m

(K. E.)f = 12{1 \over 2} m2{m \over 2} . (20)2 + 12{1 \over 2} . m2{m \over 2} (60)2 = 1000 m

| Δ\Delta K. E. | = | 1000m −- 800 m | = 200 m

ΔK.E.(K.E.)i=200m800m=14=x4{{\Delta K.E.} \over {{{(K.E.)}_i}}} = {{200m} \over {800m}} = {1 \over 4} = {x \over 4}

x = 1

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