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Standard XII

Physics

Net Force

Two unequal m...

Question

Two unequal masses ( $m_{1}$ and $m_{2}$ ) are connected by a string which passes over a frictional pulley. If $M_{1}, M_{2}$ and the table are frictionless, the acceleration of the masses would be

\frac{M_{1} g}{M_{1} + M_{2}}

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\frac{M_{1} + M_{2}}{M_{1} g}

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\frac{M_{2} g}{M_{1} + M_{2}}

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None of these

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Solution

The correct option is A $\frac{M_{1} g}{M_{1} + M_{2}}$
For Mass $M_{2}$

$T = M_{2} a$

for $M_{1}$

$T + M_{1} a = M_{1} g$

$M_{2} a + M_{1} a = M_{1} g$

$a = \frac{M_{1} g}{M_{2} + M_{1}}$

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Two unequal masses (m1 and m2) are connected by a string which passes over a frictional pulley. If M1,M2 and the table are frictionless, the acceleration of the masses would be

The correct option is A M1gM1+M2For Mass M2T=M2afor M1T+M1a=M1gM2a+M1a=M1ga=M1gM2+M1

Two unequal masses ( $m_{1}$ and $m_{2}$ ) are connected by a string which passes over a frictional pulley. If $M_{1}, M_{2}$ and the table are frictionless, the acceleration of the masses would be

The correct option is A $\frac{M_{1} g}{M_{1} + M_{2}}$
For Mass $M_{2}$

$T = M_{2} a$

for $M_{1}$

$T + M_{1} a = M_{1} g$

$M_{2} a + M_{1} a = M_{1} g$

$a = \frac{M_{1} g}{M_{2} + M_{1}}$