Two spheres of mass 10g and 100g each falls on two pans of a table balance from a height of 40cm and 10cm repectively. If both are brought to rest in 0.1s, determine the force exerted by each sphere on the pans

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Solution

The data provided by the question are:
$m_{1} = 10 g m = 10 \times 10^{- 3} k g = 10^{- 2} k g m_{2} = 100 g m = 100 \times 10^{- 3} k g = 10^{- 1} k g h_{1} = 40 c m = 40 \times 10^{- 2} m h_{2} = 10 c m = 10 \times 10^{- 2} m t = 0.1 s$
When both the spheres touches the pan then their respective velocities will be given as,
$v_{1} = \sqrt{2 g h_{1}} = \sqrt{2 \times 9.8 m / s^{2} \times 40 \times 10^{- 2} m} = \sqrt{7.84 m^{2} / s^{2}} = 2.8 m / s a n d v_{2} = \sqrt{2 g h_{2}} = \sqrt{2 \times 9.8 m / s^{2} \times 10 \times 10^{- 2} m} = \sqrt{1.96 m^{2} / s^{2}} = 1.4 m / s$
Now using Newton's second law of motion, the force exerted by each pan will be calculated as,
$F_{1} = m_{1} a_{1} = m_{1} \times \frac{v_{1}}{t} = (10^{- 2} k g) \times \frac{2.8 m / s}{0.1 s} = 0.28 N a n d F_{2} = m_{2} a_{2} = m_{2} \times \frac{v_{2}}{t} = (10^{- 1} k g) \times \frac{1.4 m / s}{0.1 s} = 1.4 N$

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