Question

A particle of mass m1 = 3.17kg is traveling in the x-direction at a speed of v1 = 24.6m/s and collides with another particle of mass m2 = 2.45kg that is initially at rest. After the collision, the first particle is traveling with a speed of 7.96m/s at an angle of 311o with respect to the x-axis. (a) What is the magnitude of the final velocity of the second particle? (b) What angle does the trajectory of the second particle make with the x-axis?

Answer 1

Initial momentum of m1 along x-axis = m1*v1x = 3.17*24.6 = 27.77 kg m/sec
Initial momentum of m1 along y-axis = 0
Initial momentum of m2 =0

Final momentum of m1 along x-axis = m1 v1 cos 311 = 3.17*7.96*0.66 = 16.55 kg m/sec
Final momentum of m1 along y-axis = m1 v1 sin 311 = 3.17*7.96*(-0.75 = -19.04 kg m/sec
Let final velocity of m2 along x-axis be v2x and along y axis be v2y
Final momentum of m2 along x-axis = m2 v2x = 2.45 v2x kg m/sec
Applying conservation of momentum of masses along x-axis, we get
27.77 = 16.55 + 2.45 v2x
v2x = 4.6 m/s
Final momentum of m2 along y-axis = m2 v2y = 2.45 v2y
Applying conservation of momentum of masses along y-axis, we get
0 = -19.04 + 2.45 v2y
v2y = 7.8 m/s

a) magnitude of final velocity of m2 = (v2x² + v2y²)^1/2
                                                      = (4.6²+7.8²)^1/2 = 9 m/s

b) Angle made by trajectory of m2 with x-axis,
= tan^-1(v2y/v2x) = tan^-1(7.8/4.6) = 59.5 deg