How much momentum will an object of mass $10$ kg transfer to the floor, if it falls from the height of $5$ m. $(g = 10 m / s^{2})$

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Solution

Velocity when it reaches the ground $= \sqrt{(2 g H)} = \sqrt{(2 g H)} = 10 \frac{m}{s}$

Momentum when it touches the floor $= 10 k g \times 10 \frac{m}{s} = 100 \frac{k g m}{s}$

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How much momentum will an object of mass 10kg transfer to the floor, if it falls from the height of 5m. (g=10m/s2)

Velocity when it reaches the ground=√(2gH)=√(2gH)=10msMomentum when it touches the floor=10kg×10ms=100kgms

How much momentum will an object of mass $10$ kg transfer to the floor, if it falls from the height of $5$ m. $(g = 10 m / s^{2})$

Velocity when it reaches the ground $= \sqrt{(2 g H)} = \sqrt{(2 g H)} = 10 \frac{m}{s}$

Momentum when it touches the floor $= 10 k g \times 10 \frac{m}{s} = 100 \frac{k g m}{s}$