A girl riding a bicycle along a straight road with a speed of ${m s}^{- 1}$ throws a stone of mass $0.5 k g$ which has a speed of $15 {m s}^{- 1}$ with respect to the ground along with her direction of motion. The mass of the girl and bicycle is $50 k g$ . Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so?

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Solution

Yes, due to the principle of conservation of momentum.

Initial momentum = $50.5 \times 5$ $k g$ $m$ $s^{- 1}$

Final momentum = $(50 V + 0.5 \times 15)$ $k g$ $m$ $s^{- 1}$

$P_{i} = P_{f}$

$V = 4.9 m s^{- 1}$ ,

change in speed = $5 - 4.9 . = 0.1 m s^{- 1}$

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Yes, due to the principle of conservation of momentum. Initial momentum = 50.5×5 kg m s–1 Final momentum = (50V+0.5×15) kg m s–1 Pi=PfV=4.9ms−1, change in speed = 5−4.9.=0.1ms–1

Yes, due to the principle of conservation of momentum.

Initial momentum = $50.5 \times 5$ $k g$ $m$ $s^{- 1}$

Final momentum = $(50 V + 0.5 \times 15)$ $k g$ $m$ $s^{- 1}$

$P_{i} = P_{f}$

$V = 4.9 m s^{- 1}$ ,

change in speed = $5 - 4.9 . = 0.1 m s^{- 1}$