Question

Problem 4 An object's position function is given by x1(t) = 5 + 10t (with 2.1 in meters if t is in seconds). A second object's position function is 2(t) 5-6t. (a) If the first object's mass is 1/3 the mass of the second one, what is the position of the system's center of mass as a function of time? (b) Under the same assumption, what is the velocity of the system's center of mass?

Answer 1

postion function = x_1 (t) = 5 + 10t and x_2 (t) = 5 - 6t m_1 = 1/3 m_2 postion of centre of mass = X_cm (t) = m1 x1 + m2 x2/m1 + m2 = m1(5 + 10t) + 3m1(5 - 6t)/m1 + 3m1 = 5 - 0.5t b) velocity of m_1 = v_1 = dx_1 (t)/dt = 10: velocity of m_2 = v_2 = dx_2 (t)/dt = -6 velocity of centre of mass = V_cm = m1 v1 + m2 v2/m1 + m2 = 10m1 - 18m1/4m1 = -2.0 m/s