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.

A
1 : 2
B
2 : 3
C
3 : 2
D
5 : 3

Solution

## The correct option is B 2 : 3$$f_1 = (2m)(acc.)$$                       $$= 2m \times \dfrac{P}{5m} = \dfrac{2}{5} P$$                 $$f_2 = (3m) (acc.)$$                      $$= 3m \times \dfrac{P}{5m} = \dfrac{3}{5} P$$ $$\dfrac{f_1}{f_2} = \dfrac{ 2 P/5}{3P/5} = \dfrac{2}{3}$$ Physics

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