Two spheres of mass $10 g$ and $100 g$ each falls on the two pans of a table balance from a height of $40 c m$ and $10 c m$ , respectively. If both are brought to rest in $0.1 s e c o n d$ , determine the force exerted by each sphere on the pans.

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Solution

Step 1: Given data
$s p h e r e o f m a s s (m_{1}) = 10 g = 0.01 k g s p h e r e o f m a s s (m_{2}) = 100 g = 0.1 k g h e i g h t (h_{1}) = 40 c m = 0.4 m h e i g h t (h_{2}) = 10 c m = 0.1 m t i m e (t) = 0.1 s$
Step 2: Finding the force exerted in the case of the first sphere:
Here, $m_{1} = 10 g = 0.01 k g$
Velocity of the sphere just before hitting the pan,
$v^{2} = u^{2} + 2 a s \Rightarrow v^{2} = u^{2} + 2 g h_{1}$ .
(Let $g = 10 m s^{- 2}$ ) and $I n i t i a l v e l o c i t y = u$ .
$v^{2} = 0 + 2 \times 10 \times 0.4 v^{2} = 8 v = 2 \sqrt{2} m s^{- 1}$
Force exerted by the first sphere,
$F = m a F = m_{1} (\frac{v - u}{t}) F = 0.01 (\frac{2 \sqrt{2} - 0}{0.1}) F = 0.2 \sqrt{2} N$ .
Step 3: Finding the force exerted in the case of the second sphere:
Here, $m_{2} = 100 g = 0.1 k g$
Velocity of the sphere just before hitting the pan,
$v^{2} = u^{2} + 2 g h_{2}$
$v^{2} = 0 + 2 \times 10 \times 0.1 v^{2} = 2 v = \sqrt{2} m s^{- 1}$
Force exerted by the second sphere,
$F^{'} = m a F^{'} = m_{2} (\frac{v - u}{t}) F^{'} = 0.1 (\frac{\sqrt{2} - 0}{0.1}) F^{'} = \sqrt{2} N$ .
Hence, the force exerted by the sphere ( $m_{1}$ ) is $0.2 \sqrt{2} N$ and sphere ( $m_{2}$ ) is $\sqrt{2} N$ .

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

Two spheres of mass $10 g$ and $100 g$ each falls on the two pans of a table balance from a height of $40 c m$ and $10 c m$ , respectively. If both are brought to rest in $0.1 s e c o n d$ , determine the force exerted by each sphere on the pans.