Physics - Center of Mass Question with Solution | TestHub
Two friends A and B (each weighing 40 kg) are sitting in a frictionless well having vertical circular smooth wall. They subtend angle 90° at the centre of the well. A rolls a ball of mass 40 kg in the clockwise direction on the platform base towards B along the wall of the well, which B catches. Then similarly B rolls the ball towards A anticlockwise and A catches it. The ball has a fixed speed of 5 m/s. (Neglect catching and throwing time).
Find the total impulse by the wall of well on whole system by the time ball travels from A to B
Options:
Answer:
Solution:
System = A + B + ball. The only external horizontal force on the system is the normal reaction of the smooth vertical wall, so the wall's impulse equals the change in linear momentum of the whole system. A rolls the ball at 5 m/s relative to himself; since A and the ball have equal mass (40 kg), momentum conservation gives the ball 2.5 m/s clockwise and A 2.5 m/s anticlockwise, both sliding along the wall. B stays at rest until the ball arrives. Let omega = 2.5/R. At time t the ball is at angle 180 - wt and A at 180 + wt, so p_ball = 100(sin wt, cos wt) and p_A = 100(sin wt, -cos wt) components add to P(t) = (200 sin wt, 0). The ball reaches B after covering 90 degrees, i.e. wt = 90 degrees, giving P = 200 kg-m/s along the x-direction. Hence the total impulse by the wall on the whole system = 200 kg-m/s. Answer: (A).
