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Standard XII

Physics

Center of Mass as an Average Point

Two blocks ma...

Question

Two blocks masses '3m' and '2m' are in contact on a smooth table. A force P is first applied horizontally on block of mass '3m' and then on mass '2m'. The contact force between the two block sin the two cases are in the ratio.

1 : 2

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2 : 3

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3 : 2

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5 : 3

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Solution

The correct option is B 2 : 3
$f_{1} = (2 m) (a c c .)$

$= 2 m \times \frac{P}{5 m} = \frac{2}{5} P$

$f_{2} = (3 m) (a c c .)$
$= 3 m \times \frac{P}{5 m} = \frac{3}{5} P$
$\frac{f_{1}}{f_{2}} = \frac{2 P / 5}{3 P / 5} = \frac{2}{3}$

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Two blocks masses '3m' and '2m' are in contact on a smooth table. A force P is first applied horizontally on block of mass '3m' and then on mass '2m'. The contact force between the two block sin the two cases are in the ratio.

The correct option is B 2 : 3f1=(2m)(acc.) =2m×P5m=25P f2=(3m)(acc.) =3m×P5m=35P f1f2=2P/53P/5=23

The correct option is B 2 : 3
$f_{1} = (2 m) (a c c .)$

$= 2 m \times \frac{P}{5 m} = \frac{2}{5} P$

$f_{2} = (3 m) (a c c .)$
$= 3 m \times \frac{P}{5 m} = \frac{3}{5} P$
$\frac{f_{1}}{f_{2}} = \frac{2 P / 5}{3 P / 5} = \frac{2}{3}$