Referring to the figure above, answer the following questions. A Particle of mas

Question

Referring to the figure above, answer the following questions. A Particle of mass m1 = 1,000 kg moves at speed v1 = 0.500 m/s. It collides with a particle of mass m2 = 2,000 kg rest What is the total momentum of the system in the x direction before the collision? Answer: __________ What is the total momentum of the system in the y direction before the collision? Answer: __________ After the collision, m1 moves with speed v3 at an angle theta2 = 315.0 degree with respect to the x axis, and m2 moves with v2 at an angle theta

Explanation / Answer

Mass of the particles m1 = 1 kg and m2 = 2 kg speed of the particle m1 is v1 = 0.5 m/s speed of the particle m2 is zero a) The total momentum of the system in X direction is Pix = m1v1 + m2(0) = 0.5 kgm/s b) The total momentum of the system in Y direction is Piy = m1(0) + m2(0) = 0 After the collision m1 travells at speed v3 making angle 1 = 315o with X axis and m2 travells at speed v2 making angle 2 = 30o with X axis The X component of velocity of m1 is v3x = v3cos1 The Y component of velocity of m1 is v3y = v3sin1 The X component of velocity of m2 is v2x = v2cos2 The Y component of velocity of m2 is v2y = v2sin2 c) The total momentum of the system in X direction is Pfx = m1v3cos1 + m2v2cos2 Pfx = 0.7071v3 + 1.732v2 d) The total momentum of the system in Y direction is Pfy = m1v3sin1 + m2v2sin2 Pfy = -0.7071v3 + v2 According to law of conservation of linear momentum, the total momentum before collision is equal to the total momentum after collision In X direction: 0.5 kgm/s = 0.7071v3 + 1.732v2 In Y direction: 0 = -0.7071v3 + v2 0.7071v3 = v2 Therefore    0.5 kgm/s = 0.7071v3 + 1.732(0.7071v3 )    v3 = 0.259 m/s    v2 = 0.183 m/s The total momentum of the system in Y direction is Piy = m1(0) + m2(0) = 0 After the collision m1 travells at speed v3 making angle 1 = 315o with X axis and m2 travells at speed v2 making angle 2 = 30o with X axis The X component of velocity of m1 is v3x = v3cos1 The Y component of velocity of m1 is v3y = v3sin1 The X component of velocity of m2 is v2x = v2cos2 The Y component of velocity of m2 is v2y = v2sin2 The Y component of velocity of m1 is v3y = v3sin1 The X component of velocity of m2 is v2x = v2cos2 The Y component of velocity of m2 is v2y = v2sin2 The X component of velocity of m2 is v2x = v2cos2 The Y component of velocity of m2 is v2y = v2sin2 The Y component of velocity of m2 is v2y = v2sin2 c) The total momentum of the system in X direction is Pfx = m1v3cos1 + m2v2cos2 Pfx = 0.7071v3 + 1.732v2 d) The total momentum of the system in Y direction is Pfy = m1v3sin1 + m2v2sin2 Pfy = -0.7071v3 + v2 The total momentum of the system in Y direction is Pfy = m1v3sin1 + m2v2sin2 Pfy = -0.7071v3 + v2 According to law of conservation of linear momentum, the total momentum before collision is equal to the total momentum after collision In X direction: 0.5 kgm/s = 0.7071v3 + 1.732v2 In Y direction: 0 = -0.7071v3 + v2 0.7071v3 = v2 Therefore    0.5 kgm/s = 0.7071v3 + 1.732(0.7071v3 )    v3 = 0.259 m/s    v2 = 0.183 m/s    v2 = 0.183 m/s

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