Question 9.125: An atomic nucleus at rest at the origin of an xy coordinate ......
An atomic nucleus at rest at the origin of an xy coordinate system transforms into three particles. Particle 1, mass 16.7 × 10^{-27} kg, moves away from the origin at velocity (6.00 × 10^{6} m/s)\hat{i}; particle 2, mass 8.35 × 10^{-27} kg, moves away at velocity (-8.00 × 10^{6}m/s)\hat{j}. (a) In unit-vector notation, what is the linear momentum of the third particle, mass 11.7 × ^{-27} kg? (b) How much kinetic energy appears in this transformation?
(a) Since the initial momentum is zero, then the final momenta must add (in the vector sense) to 0. Therefore, with SI units understood, we have
(b) Dividing by m_3=11.7 \times 10^{-27} kg and using the Pythagorean theorem we find the speed of the third particle to be v_3 = 1.03 × 10^7 m/s. The total amount of kinetic energy is \frac{1}{2}m_1 v_1^2+\frac{1}{2}m_2 v_2^2+\frac{1}{2}m_3 v_3^2=1.19 \times 10^{-12}\,J.