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Question Number 160138 by otchereabdullai@gmail.com last updated on 25/Nov/21

$$\mathrm{The}\mathrm{nucleus}\mathrm{of}\mathrm{a}\mathrm{certain}\mathrm{atom}\mathrm{has}$$$$\mathrm{mass}\mathrm{of}3.8\times {10}^{-25}\mathrm{and}\mathrm{is}\mathrm{at}\mathrm{rest}.\mathrm{The}$$$$\mathrm{nucleus}\mathrm{is}\mathrm{radioactive}\mathrm{and}\mathrm{suddenly}$$$$\mathrm{eject}\mathrm{from}\mathrm{its}\mathrm{self},\mathrm{a}\mathrm{particle}\mathrm{of}\mathrm{mass}$$$$6.6\times {10}^{-27}\mathrm{kg}\mathrm{and}\mathrm{speed}1.5\times {10}^3\mathrm{m}/\mathrm{s}.$$$$\mathrm{Find}\mathrm{the}\mathrm{recoil}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{nucleus}$$$$\mathrm{as}\mathrm{is}\mathrm{left}\mathrm{behind}.$$$$$$

Commented by mr W last updated on 25/Nov/21

$$m=totalmass=3.8\times {10}^{-25}kg$$$$m_1=massejected=6.6\times {10}^{-27}kg$$$$m_2=massremaining=m-m_1$$$$v_1=speedofmassejected=1.5\times {10}^{3}m/s$$$$v_2=recoilspeedofremainingmass$$$$m_1{v}_1+m_2{v}_2=0$$$$v_2=-\frac{m_1{v}_1}{m-m_1}=-\frac{v_1}{\frac{m}{m_1}-1}$$$$=-\frac{1.5\times {10}^3}{\frac{380}{6.6}-1}=-26.5m/s$$

Commented by otchereabdullai@gmail.com last updated on 25/Nov/21

$$\mathrm{may}\mathrm{Almity}\mathrm{Allah}\mathrm{bless}\mathrm{you}\mathrm{and}\mathrm{bless}$$$$\mathrm{you}\mathrm{again}\mathrm{prof}!$$

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