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

Correct answer: C. 0.67 ~J aligned & m_1 u_1+m_2 u_2=(m_1+m_2) v & v=2 / 3 ~m/s & Energy loss =(1)/(2)(0.5) ×(2)^2-(1)/(2)(1.5) × ((2)/(3))^2=0.67 ~J aligned

JEE Main 2008

Exam: JEE Main
Year: 2008
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

JEE Main 2008 · Physics · Center of Mass, Momentum and Collision

A block of mass 0.50 ~kg is moving with a speed of 2.00 ~m / s on a...

A block of mass

0.50 \mathrm{~kg}

is moving with a speed of

2.00 \mathrm{~m} / \mathrm{s}

on a smooth surface. It strikes another mass of

1.00 \mathrm{~kg}

and then they move together as a single body. The energy loss during the collision is

Options

A $0.16 \mathrm{~J}$
B $1.00 \mathrm{~J}$
C $0.67 \mathrm{~J}$
D $0.34 \mathrm{~J}$

Correct answer

C. $0.67 \mathrm{~J}$

Step-by-step solution

\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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