Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

31 of 130

Question 31 of 130

A block of mass 0.50 kg0.50 \mathrm{~kg} is moving with a speed of 2.00 m/s2.00 \mathrm{~m} / \mathrm{s} on a smooth surface. It strikes another mass of 1.00 kg1.00 \mathrm{~kg} and then they move together as a single body. The energy loss during the collision is

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Answer: (C) 0.67 J0.67 \mathrm{~J}

m1u1+m2u2=(m1+m2)vv=2/3 m/s Energy loss =12(0.5)×(2)2−12(1.5)×(23)2=0.67 J\begin{aligned} & m_1 u_1+m_2 u_2=\left(m_1+m_2\right) v \\ & v=2 / 3 \mathrm{~m} / \mathrm{s} \\ & \text { Energy loss }=\frac{1}{2}(0.5) \times(2)^2-\frac{1}{2}(1.5) \times\left(\frac{2}{3}\right)^2=0.67 \mathrm{~J} \end{aligned}

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