An evil doer has stolen the unstable atomic nuclear mass (17.0 x 10-27 kg) from

Question

An evil doer has stolen the unstable atomic nuclear mass (17.0 x 10-27 kg) from the unstable nuclear mass facility. While waiting at a stop light the unthinkable happens (actually quite expected and predictable up to a probability) and the mass disintegrates into three particles. One of the particles, of mass 5.00 x 10-27 kg, moves in the y-direction with a speed of 6.00 x 106 m/s. Another particle, of mass 8.40 x 10-27 kg, moves in the x-direction with a speed of 4.00 x 106 m/s. If he had been driving with a speed of 30 m/s when the disintegration occured, how would this have changed your answer to part (a)? Part A for reference : Find the magnitude and direction of the velocity of the third particle

Explanation / Answer

Momentum's conservation: m1 v1 + m2 v2 + m3 v3 = 0 (vectors) X axis : m2 v2 + m3 v3x = 0 v3x = - m2 v2/m3 Y axis : m1 v1 + m3 v3y = 0 v3y = - m1 v1/m3 m3 = m - (m1 + m2) = 1.92*10^-26 - (5.02 + 8.5)*10^-27 = 5.68*10^-27 kg So : v3x = - 5.98*10^6 m/s ; v3y = - 5.30*10^6 m/s (a) The vector v3 = - (5.98 i + 5.30 j)*10^6 m/s (b) The total kinetic energy is: Ek = (1/2)(m1 v1^2 + m2 v2^ + m3 v3^2) Ek = 3.40 * 10^13 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.