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Block A has a mass of 3 kg and is sliding on a rough horizontal surface with a velocity $v_{A} = 2 {ms}^{- 1}$ when it makes a direct collision with block B, which has a mass of 2 kg and is originally at rest. If the collision is perfectly elastic, calculate the distance, in cm, between the blocks when they stop sliding. The coefficient of kinetic friction between the blocks and the plane is $μ_{k} = 0.4$ and given $g = 10 {ms}^{- 2}$ .

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answer is 70.

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Detailed Solution

$v_{A}^{'} = (\frac{m_{A} - m_{B}}{m_{A} + m_{B}}) v_{A} = (\frac{3 - 2}{3 + 2}) (2) = 0.4 {ms}^{- 1}$

$v_{B}^{'} = (\frac{2 m_{A}}{m_{A} + m_{B}}) v_{B} = (\frac{6}{3 + 2}) (2) = 2.4 {ms}^{- 1}$

Retardation, of both the blocks will be

$a = μ_{k} g = 0.4 \times 10 = 4 {ms}^{- 2}$

$s_{A} = \frac{v_{A}^{2}}{2 a} and s_{B} = \frac{v_{B}^{' 2}}{2 a}$

Therefore, the desired distance is,

$d = s_{B} - s_{A} = \frac{v_{B}^{, 2} - v_{A}^{2}}{2 a} = \frac{(2.4)^{2} - (0.4)^{2}}{8}$

$d = \frac{5.6}{8} = 0.7 m = 70 cm$

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