Answer:
a The velocity of the third particle is [tex]v_3 = (-15.629*10^{6}i -14.45*10^{6}j)\ m/s[/tex]
b The total kinetic energy increase [tex]E = 6.50 *10^{-13} J[/tex]
Explanation:
To solve this question we are going to be using the concept of conservation of momentum which is mathematically represented as
[tex]m_1v_1 +m_2v_2 +m_3v_3 = 0[/tex]
From the question
Mass of unstable atomic nucleus [tex]m_0= 1.58 *10^{-26} kg[/tex]
Mass of x-axis particle [tex]m_1 = 8.44* 10^{-27} kg[/tex]
Velocity of x-axis particle [tex]v_ 1 = 4.00* 10^6 m/s[/tex]
Mass of y-axis particle [tex]m_2 =5.20* 10^{-27} kg[/tex]
Velocity of y-axis particle [tex]v_2 = 6.00 *10^6 m/s[/tex]
Since the initial mass is known we can obtain the mass of the third particle
Using the conservation of mass mathematical expression
[tex]m_3 = m_0 - (m_1 +m_2)[/tex]
[tex]m_3 = 1.58 *10^{-26} - ( 8.44* 10^{-27} + 5.20* 10^-27)[/tex]
[tex]= 2.16 *10 ^{-27}kg[/tex]
Using the mathematical expression above for conservation of momentum
[tex](8.44 * 10^{-27})( 4.00* 10^6i) + (5.20 *10^{-27})(6.00* 10^6j) +(2.16*0^{-27})v_3 =0[/tex]
[tex]v_3 = (-15.629*10^{6}i -14.45*10^{6}j)\ m/s[/tex]
The above is the vector solution now to obtain the value in numeric form we evaluate the magnitude of the vector
[tex]Mag = \sqrt{(15.629*10^6)^2 +(14.45*10^6)^2}[/tex]
[tex]= 21.285*10^6 \ m/s[/tex]
To obtain the total increase in kinetic energy which is mathematically represented as
[tex]E = \frac{1}{2}m_1v_1 + \frac{1}{2}m_2v_2 + \frac{1}{2} m_3 v_3[/tex]
[tex]E =\frac{1}{2} [(5.20*10^{-27})(6.0*0^{6})^2 + (8.44*10^{-27})(4.00*10^6)^2+\\\\(2.16*10^{-27})(21.285*10^6)^2][/tex]
[tex]= 6.50 *10^{-13} J[/tex]
