physic 難題 二

Q11. Let a be the angle at which v makes with the original direction of motion of the mass with 6 m/s. Using conservation of momentum along the original direction of mass with 6 m/s,6m = (2m).vcos(a)where m is the mass of a particlei.e. 3 = v.cos(a) Consider momentum conservation perpendicular to direction of motion of mass with 6 m/s12m = (2m).v.sin(a)i.e. 6 = v.sin(a) Dividing the two equations: tan(a) = 6/3 = 2i.e. a = 63.43 degreessubstitute the value of a into the equation 3 = v.cos(a)v = 3/cos(63.43) m/s = 6.71 m/sThe answer is option C Q13. Statements B and C are true. Q15. Using conservation of angular momentum,20w = 5w’where w and w’ are the angular velocities before and after the skater lifted his armsw’ = 4w Original rotational kinetic energy = (1/2).(20)w^2 = 10w^2Final rotational kinetic energy = (1/2).(5)w’^2 = (5/2).(4w)^2 = 40w^2 Hence, the rotational kinetic energy increase by a factor of= 40w^2/10w^2 = 4 The answer is option F