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Question Number 183036 by Mastermind last updated on 18/Dec/22

$$\mathrm{A}\mathrm{bullet}\mathrm{of}\mathrm{mass}1\mathrm{kg}\mathrm{is}\mathrm{fired}\mathrm{and}\mathrm{get}$$$$\mathrm{embedded}\mathrm{into}\mathrm{a}\mathrm{block}\mathrm{of}\mathrm{wood}\mathrm{of}$$$$\mathrm{mass}1\mathrm{kg}\mathrm{initially}\mathrm{at}\mathrm{rest}\mathrm{the}\mathrm{velocity}$$$$\mathrm{of}\mathrm{the}\mathrm{bullet}\mathrm{before}\mathrm{collision}\mathrm{is}90\mathrm{m}/\mathrm{s}$$$$1)\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{velocity}\mathrm{of}\mathrm{the}\mathrm{system}$$$$\mathrm{after}\mathrm{collision}?$$$$2)\mathrm{Calculate}\mathrm{the}\mathrm{kinetic}\mathrm{energy}\mathrm{before}$$$$\mathrm{and}\mathrm{after}\mathrm{the}\mathrm{collision}.$$$$3)\mathrm{How}\mathrm{much}\mathrm{energy}\mathrm{is}\mathrm{lost}\mathrm{in}\mathrm{collision}?$$

Answered by mr W last updated on 19/Dec/22

$$1)$$$$\frac{1\times 90}{1+1}=45m/s$$$$2)$$$${KE}_1=\frac{1\times {90}^2}{2}$$$${KE}_2=\frac{(1+1)\times {45}^2}{2}$$$$3)$$$$\mathrm{\Delta }KE=\frac{1\times {90}^2}{2}-\frac{(1+1)\times {45}^2}{2}$$

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