A body of mass {m₁} moving with uniform velocity of 40 ,m/s collides with another mass {m₂} at r

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5.Work, Energy, Power and Collision

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A body of mass ${m_1}$ moving with uniform velocity of $40 \,m/s$ collides with another mass ${m_2}$ at rest and then the two together begin to move with uniform velocity of $30\, m/s$. The ratio of their masses $\frac{{{m_1}}}{{{m_2}}}$ is

A

$0.75$

B

$1.33$

C

$3$

D

$4$

Solution

(c) Initial momentum of the system = ${m_1} \times 40 + {m_2} \times 0$

Final momentum of the system = $({m_1} + {m_2}) \times 30$

By the law of conservation of momentum

${m_1} \times 40 + {m_2} \times 0 = ({m_1} + {m_2}) \times 30$

==> $40{m_1} = 30{m_1} + 30{m_2}$ ==> $10{m_1} = 30{m_2}$=$\frac{{{m_1}}}{{{m_2}}} = 3$

Standard 11

Physics

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