A ball whose mass is $100 g$ is dropped from a height of $2 m$ from the floor. It rebounds vertically up wards after colliding with the floor to a height $1.5 m$ . Find $(a)$ the momentum of the ball before and after colliding with the floor, $(b)$ the average force exerted by the floor on the ball. Assume that collision lasts for $10^{- 8} s$ .

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Solution

let the downward velocity be $V_{1}$ and upward velocity be $V_{2}$
now $V_{1} = \sqrt{2 g h_{1}} = \sqrt{2 \times 9.8 \times 2} = \sqrt{4 \times 9.8} = 2 \sqrt{9.8}$
momentum before collision = $m V_{1}$ = $100 \times 10^{- 3} \times 2 \sqrt{9.8}$
= $0.626 \frac{k g m}{s}$

after collision $V_{2} = \sqrt{2 g h_{2}} = \sqrt{2 \times 9.8 \times 1.5} = \sqrt{3 \times 9.8}$
momentum after collision = $- m V_{2}$ = $- 100 \times 10^{- 3} \sqrt{3 \times 9.8}$
= $- 0.542 \frac{k g m}{s}$ direction is reverse therefore negative sign assigned
$△ P = P_{f} - P_{i}$ = -0.542-0.626 = -1.168
$F = \frac{△ P}{△ t} = 1.168 \times 10^{8} N$

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Similar questions

A ball is dropped from a height $h_{0}$ on a horizontal floor and keeps rebounding after hitting the floor. Find: (i) speed after $n$ rebounds

(ii) Total distance travelled by the ball before stopping

(iii) Height it rises after nth rebound

(iv) Total time before stopping

(coefficient of restitution is $e$ )

10 m

2.5 m

0.01 s

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