Question
0 My Notes OAsk Your 2.0 g particle moving at 5.4 m/s makes a perfectly elastic head-on collision with a resting 1.0 g object. (a) Find the speed of each particde after the collision. 2.0 g particle 1.0 g particle Textbor mVs m/s (b) Find the speed of each particle after the collision if the stationary particle has a mass of 109 2.0 g particle mys 10.0 g partidle m/s in (c) Find the final kinetic energy of the Incident 2.0 g particle in the situations described In parts (a) and (b) KE In part (a) KE in part(b) In which case does the incident partide lose more kinetic energy? O case (b) O case (a) Need Help? Tan to izing My Notes Ask Your points SerCP8 &.P060 particles. One coordinate system, disintegrates into three nucleus of mass 1.70 x 102 kg, Initially at rest at the origin of a An unstable nucieus of mass 1.70 x 10 particle, having a mass of m1 5.60 x 1027 kg moves in the pesitive y-direction with s mass m2 - 8.40 x 1027 Assume that mass is conserved in this process. Indicate the peed i 4.80x 10 mys. Ancther particle, of tain kg moves in the positive x-direction with speed v2-4.00 x 1a m/s. Find the velocity of the third partide. direction with the sign af your answer.) m/s
Explanation / Answer
a] By Conservation of Momentum
Before Vcg = 2 /3 * 5.4 m/sec
V1 before = Vcg +1/2Vcg = 3/2 Vcg
V1 after = Vcg -1/2Vcg = 1/2 Vcg = 1/3 * 5.4 m/sec = 1.8 m/s
V2 before =Vcg - Vcg = 0
V2 after = Vcg + Vcg = 2 Vcg = 4/3 *5.4 m/s = 7.2 m/s
Speed of 2g particle after collision = 1.8 m/s
Speed of 1g particle after collision = 7.2 m/s
b] Vcg = 1/6 * 5.4 m/sec
V1 = Vcg - 5 Vcg = -4Vcg = -4/6 *5.4 m/s = - 3.6 m/s
V2 = Vcg +Vcg = 2 Vcg = 2/6 * 5.4 m/sec = 1.8 m/s
Speed of 2g object = -3.6 m/s
Speed of 1g object = 1.8 m/s
c] For 2g object,
KE before collision = 0.5 *2*10^-3 * 5.4^2 = 0.02916 J
KE after collision = 0.5*2*!0^-3*1.8^2 = 0.00324 J
For 1g object
KE before collision = 0
KE after collision = 0.5*1*10^-3*7.2^2 = 0.02592
d] In case a
b. KE in = 0.0656, KE out = (2/3)^2 * 0.0656 = 0.02961 N