A force of 20 N is applied on two balls of masses $m_{1} = 2 k g$ , and $m_{2} = 4$ kg individually. Find the ratio of accelerations of balls 1 and 2?

1:1

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Solution

The correct option is C 2:1
According to newton’s second law,
$F = m a$
where terms have their usual meanings.

$a = \frac{F}{m}$
$⟹ a \propto \frac{1}{m}$
$⟹ \frac{a_{1}}{a_{2}} = \frac{m_{2}}{m_{1}}$
Putting values, we get:
$\frac{a_{1}}{a_{2}} = \frac{4}{2}$
$a_{1} : a_{2} = 2 : 1$

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Q. Two blocks of mass

4 k g

and

2 k g

are placed side by side on a smooth horizontal surface as shown in the figure. A horizontal force of

20 N

is applied on

4 k g

block. Find the acceleration of each block.

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A force of 20 N is applied on two balls of masses m1=2 kg, and m2=4 kg individually. Find the ratio of accelerations of balls 1 and 2?

The correct option is C 2:1According to newton’s second law, F=ma where terms have their usual meanings. a=Fm ⟹a∝1m ⟹a1a2=m2m1 Putting values, we get: a1a2=42 a1:a2=2:1

A force of 20 N is applied on two balls of masses $m_{1} = 2 k g$ , and $m_{2} = 4$ kg individually. Find the ratio of accelerations of balls 1 and 2?

The correct option is C 2:1
According to newton’s second law,
$F = m a$
where terms have their usual meanings.

$a = \frac{F}{m}$
$⟹ a \propto \frac{1}{m}$
$⟹ \frac{a_{1}}{a_{2}} = \frac{m_{2}}{m_{1}}$
Putting values, we get:
$\frac{a_{1}}{a_{2}} = \frac{4}{2}$
$a_{1} : a_{2} = 2 : 1$