Question

An unstable atomic nucleus of mass 1.58 10-26 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.20 10-27 kg, moves in the y direction with a speed of 6.00 106 m/s. Another particle, of mass 8.44 10-27 kg, moves in the x direction with a speed of 4.00 106 m/s.
(a) Find the velocity of the third particle.
i + j m/s
(b) Find the total kinetic energy increase in the process.
J

Answer 1

Answer:

a The velocity of the third particle is $v_3 = (-15.629*10^{6}i -14.45*10^{6}j)\ m/s$

b The total kinetic energy increase $E = 6.50 *10^{-13} J$

Explanation:

To solve this question we are going to be using the concept of conservation of momentum which is mathematically represented as

            $m_1v_1 +m_2v_2 +m_3v_3 = 0$

From the question

        Mass of  unstable atomic nucleus $m_0= 1.58 *10^{-26} kg$

        Mass of  x-axis particle $m_1 = 8.44* 10^{-27} kg$

        Velocity of x-axis particle $v_ 1 = 4.00* 10^6 m/s$

         Mass of  y-axis particle  $m_2 =5.20* 10^{-27} kg$

            Velocity of y-axis particle  $v_2 = 6.00 *10^6 m/s$

 Since the initial mass is known we can obtain the mass of the third particle  

 Using the conservation of mass mathematical expression

                $m_3 = m_0 - (m_1 +m_2)$

                  $m_3 = 1.58 *10^{-26} - ( 8.44* 10^{-27} + 5.20* 10^-27)$

                        $= 2.16 *10 ^{-27}kg$

Using the mathematical expression above for conservation of momentum

               $(8.44 * 10^{-27})( 4.00* 10^6i) + (5.20 *10^{-27})(6.00* 10^6j) +(2.16*0^{-27})v_3 =0$

            $v_3 = (-15.629*10^{6}i -14.45*10^{6}j)\ m/s$

The above is the vector solution now to obtain the value in numeric form we evaluate the magnitude of the vector

         $Mag = \sqrt{(15.629*10^6)^2 +(14.45*10^6)^2}$

                $= 21.285*10^6 \ m/s$

To obtain the total increase in  kinetic energy which is mathematically represented as

         $E = \frac{1}{2}m_1v_1 + \frac{1}{2}m_2v_2 + \frac{1}{2} m_3 v_3$

         $E =\frac{1}{2} [(5.20*10^{-27})(6.0*0^{6})^2 + (8.44*10^{-27})(4.00*10^6)^2+\\\\(2.16*10^{-27})(21.285*10^6)^2]$

              $= 6.50 *10^{-13} J$