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Question(s): 1. The physics of curling stone collisions and the conservation of momentum. Two identical curling stones of mass 19.5 kg collide. The stone m; is moving at 5.00 m/s to the right towards a stationary stone of mass m,. The collision is a glancing collision. After the collision, one of the stones m1 moves at 3.50 m/s at an angle of 40.0 below the horizontal. a) What is the magnitude of the x-component momentum of m, after the collision? b) What is the magnitude of the y-component momentum of m, after the collision? c) What is the magnitude of the total momentum of m, after the collision? d) What is the direction of the total momentum of m, after the collision? e) What is the magnitude of the velocity of m, after the collision? f) What is the direction of the velocity of m, after the collision? Before the Collision After the Collision Vi =5.00 m/s vy; = 0.00 m/s m; =195 kg m;=19.5 kg "'-.g_t'o.nv my =19.5 kg Vir= 3.50 m/s 2. The physics of radioactive decay and the conservation of momentum. A nucleus, initially at rest, decays radioactively. In the process, it emits an electron horizontally to the east, with momentum 9.00 x 102 kg . m/s and a "neutrino" horizontally to the south, with a momentum 4.80 x 102! kg . m/s. The mass of the residual nucleus is 3.60 x 102 kg. a) What are the x-component and the y-components of momentum of the residual nucleus after the decay? b) What is the magnitude of the total momentum of the residual nucleus after the decay? ) What is the direction of the total momentum of the residual nucleus after the decay? d) What is the magnitude and velocity of the residual nucleus after the decay? After the Decay Pr=900x10%" kg - m/s electron neutrino p:=4.80x 107 kg - m/s ;f Collision in Two Dimensions: A Problem The diagram below shows two balls moving towards each other and about to collide elastically. The mass m4=0.200 kg is moving upwards at 2.75 m/s and 20.0 as shown in the diagram below. The mass m5 = 0.086 kg is moving downwards at 4.07 m/s and 10.0 as shown below. After the collision, the mass m, is moving at 3.37 m/s and 30.0 above the horizontal. We set out to determine the magnitude and direction of the momentum of my after the collision. Before the Collision After the Collision _ v, =3.37 m/s m; = 0.086 kg m; = 0.086 kg v, = 4.07 m/s v 100 3000 1 2000 ./'5, =275 m/s my = 0.200 kg In a two-dimensional collision, the total initial momentum is equal to the total final momentum, just as was the case in the one-dimensional collision. Also, in a two- dimensional collision, the momentum is conserved is both the x-direction and the y-direction separately. Therefore, the total initial momentum before the collision in the x-direction, py;, is equal to the total final momentum after the collision in the x-direction, py, so that Pxi = Pxf Similarly, the total initial momentum before the collision in the y-direction, py;, is equal to the total final momentum after the collision in the y-direction, pys, so that py\\ = pyf In determining the final momentum of my, it will be necessary to determine the momentum of the mass after the collision by using the principles of the conservation of momentum in two dimensions. The x-component of the final momentum, P1xf, for m1 is 0.617 kg.m/s, and the y-component of the final momentum, P1yf, for this mass is -0.018 kg.m/s. A of these components is as follows. P1xf Plyf Pif 7. The magnitude of the final momentum, pif, for m, is P1f = Pixf + Piye (0.611 kg . m/s) + (-0.018 kg . m/s) = 0.611 kg.m/s 8. The angle of this momentum vector is o = tan Plys Pixf 0.018 kg . m/s = tan 0.611 kg . m/s = 1.69 below the horizontal 9. The magnitude of the final velocity is Vif = Pit = 0.611 kg . m/s 0.200 kg = 3.06 m/s 10. The angle of the velcity vector is the same as the angle of the momentum vector. O = 1.69 below the horizontal.Determining the Final Momentum and Final Velocity The x-component of the final momentum, P3xf, for m, is -0.260 kg.m/s, and the y-component of the final momentum, Payf, for this mass is 0.451 kg.m/s. A sketch of these components is as follows. P3yf P3f A P3xf 7. The magnitude of the final momentum, p3f, for m3 is P31 Pixf + Pay = (-0.260 kg . m/s) + (0.451 kg . m/s) = 0.521 kg.m/s 8. The angle of this momentum vector is o= tan Payf P3xf = tan" 0.451 kg . m/s 0.260 kg . m/s = 60.0 north of west 9. The magnitude of the final velocity is P3f V3f 0.521 kg . m/s 0.400 kg = 1.30 m/s 10. The angle of the velocity vector is the same as the angle of the momentum vector. - = 60.0' north of west.Summary In this lesson, we studied collisions and explosions in two dimensions. In a two-dimensional collision or explosion, the total initial momentum is equal to the total final momentum. The total initial momentum before the collision (or explosion) in the x-direction, pxi, is equal to the total final momentum in the x-direction, Pxf, so that Pxi = Pxf. The total initial momentum before the collision (or explosion) in the y-direction, pyi, is equal to the total final momentum in the y-direction, Pyf, so that Pyi = Pyf. To determine the momentum and velocity of a mass m, after the collision, we can follow the following steps. 1. Determine the total initial x-component of the momentum before the collision (explosion): Pxi 2. Determine the total final x-component of the momentum after the collision (explosion): Pxf 3. Determine the final x-component of the momentum of m, after the collision (explosion): P1xf 4. Determine the total initial y-component of the momentum before the collision (explosion): Pyi 5. Determine the total final y-component of the momentum after the collision (explosion): Pxf 6. Determine the final y-component of the momentum of m, after the collision (explosion): plyf 7. The magnitude of the final momentum, p1f, for m, is pif = Pixi + Pyt 8. The angle of this momentum vector is @ = tan - Plyf Pixf 9. The magnitude of the final velocity is V1 = Pit my 10. The angle of the velocity vector is the same as the angle of the momentum vector