Momentum and CollisionMedium
Question
A ball of mass 0.5 kg is dropped from a height of 40 m . The ball hits the ground and rises to a height of 10 m . The impulse imparted to the ball during its collision with the ground is (Take $g = 9.8\text{ }m/s^{2}$ )
Options
A.21 NS
B.7 NS
C.0
D.84 NS
Solution
$${v_{1} = \sqrt{2gh_{1}} }{= \sqrt{2 \times 9.8 \times 40} }{v_{1} = \sqrt{784} = 28\text{ }m{\text{ }s}^{- 1} }$$and $v_{2} = \sqrt{2gh_{2}} = \sqrt{2 \times 9.8 \times 10}$
$$= \sqrt{196} = 14\text{ }m{\text{ }s}^{- 1} $$Impulse $= \Delta\overrightarrow{p} = m\left( {\overrightarrow{v}}_{f} - {\overrightarrow{v}}_{i} \right) = m\left( {\overrightarrow{v}}_{2} - {\overrightarrow{v}}_{1} \right)$
$$= \frac{1}{2}(14 - ( - 28)) $$$= 21$ NS
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