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The-nucleus-of-a-certain-atom-has-mass-of-3-8-10-25-and-is-at-rest-The-nucleus-is-radioactive-and-suddenly-eject-from-its-self-a-particle-of-mass-6-6-10-27-kg-and-speed-1-5-10-3-m-s-Find-the

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Question Number 160138 by otchereabdullai@gmail.com last updated on 25/Nov/21
The nucleus of a certain atom has mass of 3.8×10^(−25 ) and is at rest. The nucleus is radioactive and suddenly eject from its self, a particle of mass 6.6×10^(−27) kg and speed 1.5×10^3 m/s. Find the recoil speed of the nucleus as is left behind.
$$\mathrm{The}\:\mathrm{nucleus}\:\mathrm{of}\:\mathrm{a}\:\mathrm{certain}\:\mathrm{atom}\:\mathrm{has} \\ $$$$\mathrm{mass}\:\mathrm{of}\:\mathrm{3}.\mathrm{8}×\mathrm{10}^{−\mathrm{25}\:} \mathrm{and}\:\mathrm{is}\:\mathrm{at}\:\mathrm{rest}.\:\mathrm{The} \\ $$$$\mathrm{nucleus}\:\mathrm{is}\:\mathrm{radioactive}\:\mathrm{and}\:\mathrm{suddenly} \\ $$$$\mathrm{eject}\:\mathrm{from}\:\mathrm{its}\:\mathrm{self},\:\mathrm{a}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass} \\ $$$$\mathrm{6}.\mathrm{6}×\mathrm{10}^{−\mathrm{27}} \mathrm{kg}\:\mathrm{and}\:\mathrm{speed}\:\mathrm{1}.\mathrm{5}×\mathrm{10}^{\mathrm{3}} \mathrm{m}/\mathrm{s}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{recoil}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{nucleus} \\ $$$$\mathrm{as}\:\mathrm{is}\:\mathrm{left}\:\mathrm{behind}. \\ $$$$ \\ $$
Commented by mr W last updated on 25/Nov/21
m=total mass=3.8×10^(−25) kg m_1 =mass ejected=6.6×10^(−27) kg m_2 =mass remaining=m−m_1 v_1 =speed of mass ejected=1.5×10^3 m/s v_2 =recoil speed of remaining mass m_1 v_1 +m_2 v_2 =0 v_2 =−((m_1 v_1 )/(m−m_1 ))=−(v_1 /((m/m_1 )−1)) =−((1.5×10^3 )/(((380)/(6.6))−1))=−26.5 m/s
$${m}={total}\:{mass}=\mathrm{3}.\mathrm{8}×\mathrm{10}^{−\mathrm{25}} {kg} \\ $$$${m}_{\mathrm{1}} ={mass}\:{ejected}=\mathrm{6}.\mathrm{6}×\mathrm{10}^{−\mathrm{27}} {kg} \\ $$$${m}_{\mathrm{2}} ={mass}\:{remaining}={m}−{m}_{\mathrm{1}} \\ $$$${v}_{\mathrm{1}} ={speed}\:{of}\:{mass}\:{ejected}=\mathrm{1}.\mathrm{5}×\mathrm{10}^{\mathrm{3}} {m}/{s} \\ $$$${v}_{\mathrm{2}} ={recoil}\:{speed}\:{of}\:{remaining}\:{mass} \\ $$$${m}_{\mathrm{1}} {v}_{\mathrm{1}} +{m}_{\mathrm{2}} {v}_{\mathrm{2}} =\mathrm{0} \\ $$$${v}_{\mathrm{2}} =−\frac{{m}_{\mathrm{1}} {v}_{\mathrm{1}} }{{m}−{m}_{\mathrm{1}} }=−\frac{{v}_{\mathrm{1}} }{\frac{{m}}{{m}_{\mathrm{1}} }−\mathrm{1}} \\ $$$$\:\:\:\:=−\frac{\mathrm{1}.\mathrm{5}×\mathrm{10}^{\mathrm{3}} }{\frac{\mathrm{380}}{\mathrm{6}.\mathrm{6}}−\mathrm{1}}=−\mathrm{26}.\mathrm{5}\:{m}/{s} \\ $$
Commented by otchereabdullai@gmail.com last updated on 25/Nov/21
may Almity Allah bless you and bless you again prof!
$$\mathrm{may}\:\mathrm{Almity}\:\mathrm{Allah}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{and}\:\mathrm{bless} \\ $$$$\mathrm{you}\:\mathrm{again}\:\mathrm{prof}! \\ $$

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