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  • A girl riding a bicycle along a straight road with a speed of 5 m s^–1 throws a stone of mass 0.5 kg which has a speed of 15 m s^–1 with respect to the ground along her direction of motion.

A girl riding a bicycle along a straight road with a speed of 5 m s-1 throws a stone of mass 0.5 kg which has a speed of 15 m s-1 with respect to the ground along her direction of motion. The mass of the girl and bicycle is 50 kg. Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so?

Answers (1)

Here it is given that: m1 = 50kg & m2 = 0.5 kg

                                                u1 = 5 m/s forward & u2 = 5 m/s forward

                                                v2 = 15 m/s forward

To find: v1 = ?

By the law of conservation of momentum-

Initial momentum (i.e., girl, cycle & body) = Final momentum (girl, cycle & body)

(50 + 0.5) \times 5 = 50 \times v_{1} + 0.5 \times 15

50.5 \times 5 - 7.5 = 50v_{1}

50v_{1} = 252.5 - 7.5 = 245

Thus,v_{1} = \frac{245}{50} = 4.9 m/s

Thus, the speed of the cycle and the girl is reduced by 5 – 4.9 = 0.1 m/s.

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