If a lighter body (mass $M_{1}$ and velocity $V_{1}$ ) and a heavier body (mass $M_{2}$ and velocity $V_{2}$ ) have the same kinetic energy, then:

M_{2} V_{2} < M_{1} V_{1}

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M_{2} V_{2} = M_{1} V_{1}

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M_{2} V_{1} = M_{1} V_{1}

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M_{2} V_{2} > M_{1} V_{1}

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Solution

The correct option is B $M_{2} V_{2} > M_{1} V_{1}$
$K E_{1} = K E_{2}$
$P = m v; p^{2} / 2 m = K E => P = \sqrt{2 m K E}$
$P_{1} / P_{2} = \sqrt{m_{1} / m_{2}}$
$m 2 > m 1 => P_{2} > P_{1}$

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What is M1V1=M2V2 different from M1=M2V2/(V1+V2)and N1V1=N2V2

Column A	Column B
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b) Law of conservation of momentum	ii) $M a s s \times a c c e l e r a t i o n$
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M_{1} V_{1} = M_{2} V_{2}

, as the volume increases the molarity decreases
Because
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Q. Assertion :A mass

m_{1}

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m_{2}

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If a lighter body (mass M1 and velocity V1) and a heavier body (mass M2 and velocity V2) have the same kinetic energy, then:

The correct option is B M2V2>M1V1KE1=KE2P=mv;p2/2m=KE=>P=√2mKEP1/P2=√m1/m2m2>m1=>P2>P1

If a lighter body (mass $M_{1}$ and velocity $V_{1}$ ) and a heavier body (mass $M_{2}$ and velocity $V_{2}$ ) have the same kinetic energy, then:

The correct option is B $M_{2} V_{2} > M_{1} V_{1}$
$K E_{1} = K E_{2}$
$P = m v; p^{2} / 2 m = K E => P = \sqrt{2 m K E}$
$P_{1} / P_{2} = \sqrt{m_{1} / m_{2}}$
$m 2 > m 1 => P_{2} > P_{1}$