linear momentum (urgent)

69. (a) The speed of the large ball before striking the floor
= square-root[2g x 1.8] m/s = 6 m/s (where gis the acceleration du to gravity)
After collision with the floor, the speed of the large ball is still 6 m/s but in the upward direction.

Hence, using conservation of momentum,
0.63 x 6 + m(-6) = mv
where v is the velocity of the small ball after collision with the large ball
v = (3.78 - 6m)/m

Because the collision of the two balls is elastic, kinetic energy is conerved
(1/2).(0.63).(6^2) + (1/2)m(6^2) = (1/2)mv^2
22.68 + 36m = mv^2
22.68 + 36m = m[(3.78 - 6m)/m]^2
solve for m gives m = 0.21 kg

(b) v = (3.78-6x0.21)/0.21 m/s = 12 m/s
Hence, height reached = 12^2/2g m = 7.2 m

73(a) Using conservation of momentum in direction perpendicular to initial direction of the alpha particle.
16 x 1.2x10^5.sin(51) = 4.[Vif.sin(64)]
V1f = 4.15 x 10^5 m/s

(b) Using conservation of momentum in the initial travelling direction of the alpha
4(V1i) = 4.(4.15x10^5)cos(64) + 16.(1.2x10^5)cos(51)
V1i = 4.84 x 10^5 m/s

75. Let( 2a) be the angle between the two objects, and u be the initial velocity of each object.
Using conservation of momentum in direction along the path of the two masses after collision.

mu.cos(a) + mu.cos(a) = (2m)(u/2)
i.e. 2mu.cos(a) = 2mu/2
cos(a) = 1/2 = 0.5
a = 60 degrees
Hence, the initial angle between the velocities of the two objects is 2x60 degrees = 120 degrees