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Question Number 21490 by NECx last updated on 24/Sep/17

$$\mathrm{The}\mathrm{kinetic}\mathrm{energy}\mathrm{of}\mathrm{a}\mathrm{body}$$$$\mathrm{increases}\mathrm{by}100\mathrm{\%}.\mathrm{Find}\mathrm{the}\mathrm{\%}$$$$\mathrm{increase}\mathrm{in}\mathrm{its}\mathrm{momentum}.$$$$$$$$\mathrm{please}\mathrm{solve}\mathrm{with}\mathrm{explanations}$$$$\mathrm{where}\mathrm{necessary}.\mathrm{Thanks}.$$

Answered by ajfour last updated on 25/Sep/17

$$\frac{p_1^2}{2m}=K_1$$$$\frac{p_2^2}{2m}=K_2$$$$\frac{K_2-K_1}{K_1}\times 100=100$$$$\Rightarrow \frac{p_2^2-p_1^2}{p_1^2}\times 100=100$$$${\left(\frac{p_2}{p_1}\right)}^2-1=1\Rightarrow {\left(\frac{p_2}{p_1}\right)}^2=2$$$$\mathrm{\%}changeinmomentumis$$$$\frac{p_2-p_1}{p_1}\times 100=(\sqrt{2}-1)\times 100$$$$=41.4\mathrm{\%}.$$$$unlessthechangeisdifferential$$$$changeyoucannotdifferentiate$$$$togetthecorrectanswer.$$

Commented by NECx last updated on 25/Sep/17

$$\mathrm{Mr}\mathrm{Ajfour}\mathrm{please}\mathrm{this}\mathrm{solution}$$$$\mathrm{isnt}\mathrm{explanatory}\mathrm{to}\mathrm{me}.\mathrm{help}$$$$\mathrm{simplify}.$$

Commented by NECx last updated on 25/Sep/17

$$\mathrm{i}\mathrm{did}\mathrm{it}\mathrm{this}\mathrm{way}\mathrm{so}\mathrm{i}\mathrm{want}\mathrm{to}\mathrm{see}$$$$\mathrm{whats}\mathrm{wrong};$$$$$$$$\mathrm{let}\mathrm{k}=\mathrm{kinetic}\mathrm{energy},\mathrm{m}=\mathrm{mass}$$$$\mathrm{v}=\mathrm{velocity},\rho =\mathrm{momentum}$$$$$$$$\mathrm{k}=\frac{1}{2}{\mathrm{mv}}^2......\left(1\right)$$$$\rho =\mathrm{mv}$$$$\mathrm{hence};\mathrm{v}=\rho /\mathrm{m}......\left(2\right)$$$$$$$$\mathrm{put}\left(2\right)\mathrm{into}\left(1\right)$$$$$$$$\mathrm{k}=\frac{{\rho }^2}{2\mathrm{m}}$$$$\mathrm{\Delta }\mathrm{k}=\frac{\rho }{\mathrm{m}}\mathrm{\Delta }\rho$$$$$$$$\mathrm{since}\mathrm{\Delta }\mathrm{k}=\mathrm{k}$$$$$$$$\mathrm{\Delta }\mathrm{k}=\mathrm{k}=\frac{\rho }{\mathrm{m}}\mathrm{\Delta }\rho$$$$$$$$\frac{{\rho }^2}{2\mathrm{m}}=\frac{\rho }{\mathrm{m}}\mathrm{\Delta }\rho$$$$$$$$\therefore \frac{\mathrm{\Delta }\rho }{\rho }=\frac{1}{2}$$$$$$$$\therefore \mathrm{\%}\mathrm{\Delta }\rho =\frac{\mathrm{\Delta }\rho }{\rho }\times 100=50\mathrm{\%}$$$$$$$$$$$$\mathrm{whats}\mathrm{wrong}\mathrm{here}???????$$$$$$

Commented by Joel577 last updated on 25/Sep/17

$$K=\frac{p^2}{2m}$$$$$$$$\mathrm{Let}{\mathrm{K}}_2=2K_1$$$$\frac{K_1}{K_2}=\frac{\frac{p_1^2}{2m}}{\frac{p_2^2}{2m}}\to \frac{K_1}{2K_1}=\frac{p_1^2}{p_2^2}\to p_2^2=2p_1^2$$$$$$$$p_2=\left(\sqrt{2}\right)p_1$$$$\mathrm{\%}\mathrm{increase}=\frac{\mathrm{\Delta }p}{p_1}.100\mathrm{\%}=41.4\mathrm{\%}$$

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