Center Of Mass Momentum And Collision Question 7
Question 7 - 31 January - Shift 1
A solid sphere of mass $1 kg$ rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{-3} J$.
The speed of the centre of mass of the sphere is $cm s^{-1}$.
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Answer: (10)
Solution:
Formula: Conservation of Momentum
$ \begin{aligned} & \frac{1}{2} mv^{2}+\frac{1}{2} I \omega^{2}=7 \times 10^{-3} \\ & \frac{1}{2} mv^{2}+\frac{1}{2}(\frac{2}{5} MR^{2})(\frac{V}{R})^{2}=7 \times 10^{-3} \\ & \frac{1}{2} MV^{2}[1+\frac{2}{5}]=7 \times 10^{-3} \\ & \frac{1}{2}(1)(V^{2})(\frac{7}{5})=7 \times 10^{-3} \\ & V^{2}=10^{-2} \\ & V=10^{-1}=0.1 m / s=10 cm / s \end{aligned} $ Ans : 10
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