A body of mass m_1=4 ~kg moves at 5 i~ m / s and another body of mass m_2=2 ~kg moves at 10 i~ m / s . The kinetic energy of centre of mass is

Correct answer: C. (400)/(3) ~J v_ CM =(m_1 (d r_1)/(d t)+m_2 (d r_2)/(d t))/(m_1+m_2)= 4 × 5 i+2 × 10 i 4+2 v_ CM = 40 i 6 =(20)/(3) i The kinetic energy K=(1)/(2) m v^2 =(1)/(2) ×(4+2) × (20 × 20)/(3 × 3) =(1)/(2) × 6 × (20 × 20)/(3 × 3) K=(400)/(3) ~J

JIPMER 2012

Exam: JIPMER
Year: 2012
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

JIPMER 2012 · Physics · Center of Mass, Momentum and Collision

A body of mass m_1=4 ~kg moves at 5 i~ m / s and another body of...

A body of mass

m_1=4 \mathrm{~kg}

moves at

5 \mathrm{i~} \mathrm{m} / \mathrm{s}

and another body of mass

m_2=2 \mathrm{~kg}

moves at

10 \mathrm{i~} \mathrm{m} / \mathrm{s}

. The kinetic energy of centre of mass is

Options

A $\frac{200}{3} \mathrm{~J}$
B $\frac{500}{3} \mathrm{~J}$
C $\frac{400}{3} \mathrm{~J}$
D $\frac{800}{3} \mathrm{~J}$

Correct answer

C. $\frac{400}{3} \mathrm{~J}$

Step-by-step solution

v_{\mathrm{CM}}=\frac{m_1 \frac{d r_1}{d t}+m_2 \frac{d r_2}{d t}}{m_1+m_2}=\frac{4 \times 5 \mathbf{i}+2 \times 10 \mathbf{i}}{4+2}

v_{\mathrm{CM}}=\frac{40 \mathbf{i}}{6}=\frac{20}{3} \mathbf{i}

The kinetic energy

K=\frac{1}{2} m v^2

=\frac{1}{2} \times(4+2) \times \frac{20 \times 20}{3 \times 3}

=\frac{1}{2} \times 6 \times \frac{20 \times 20}{3 \times 3}

K=\frac{400}{3} \mathrm{~J}

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