A body of mass 1 0 3 kg moving with speed 2 m s − 1 collides with a body of mass 15 × 1 0 3 kg inelastically and sticks to it. Then loss in KE of the system will be

Loss in KE = 2 1 ( M 1 + M 2 ) M 1 M 2 ( u 1 − u 2 ) 2 = 2 1 × 16 × 1 0 3 1 0 3 × 15 × 1 0 3 × ( 2 ) 2 = 2 1 × 16 15 × 1 0 3 × 4 = 1.875 kJ

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Physics - W.E.P. Question with Solution | TestHub

PhysicsW.E.P.MiscellaneousEasy2 minPYQ_2019

PhysicsEasysingle choice

A body of mass $1 0^{3} kg$ moving with speed $2 m s^{- 1}$ collides with a body of mass $15 \times 1 0^{3} kg$ inelastically and sticks to it. Then loss in KE of the system will be

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Solution:

$Loss in KE = \frac{1}{2} \frac{M _{1} M _{2}}{( M _{1} + M _{2} )} (u_{1} - u_{2})^{2} = \frac{1}{2} \times \frac{1 0 ^{3} \times 15 \times 1 0 ^{3}}{16 \times 1 0 ^{3}} \times (2)^{2} = \frac{1}{2} \times \frac{15 \times 1 0 ^{3}}{16} \times 4 = 1.875 kJ$

Stream:AIIMSSubject:PhysicsTopic:W.E.P.Subtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2019

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