Question

Particle 1 has mass 3.42 kg and its position in meters as a function of time is given by r₁(t) = 2.00t i + 7.00t²j. Particle 2 has mass 5.00 kg and its position in meters as a function of time is given by r₂(t) = 7.00 i - 8.00t³j.

(a) At time t = 0.433 s, the center-of-mass of the two particles is located at
m/s i + m/s j
(b) At time t = 0.658 s, the velocity of the center of mass is
m/s i +   m/s j
(c) At time t = 0.826 s, the acceleration of the center of mass is
m/s²i +   m/s²j

Not sure where to start, any help would be appreciated!

Answer 1

(a) Position of the center of mass of the two particle is gien as -

r(t) = [m1*r1(t) + m2*r2(t)] / (m1 + m2)

= [3.42*(2.00ti + 7.00t^2j) + 5.0*(7.0i - 8.0t^3j)] / (3.42+5.0)

= [6.84ti + 23.94t^2j + 35i - 40t^3j] / 8.42

= [(35 + 6.84t)i + (23.94t^2 - 40t^3)j] / 8.42

= (4.16 + 0.81t)i + (2.84t^2 - 4.75t^3)j

put t = 0.433 s

So, Position of the center-of-mass at t = 0.433 s

r(0.433) = (4.16 + 0.81*0.433)i + (2.84*0.433^2 - 4.75*0.433^3)j

= (4.16 + 0.35)i + (0.53 - 0.38)j = (4.51i + 0.15j) m

(b) v(t) = dr(t) / dt = 0.81i + (2*2.84t - 3*4.75t^2)j

= 0.81i + (5.68t - 14.25t^2)j

So, v(0.658) = 0.81i + (5.68*0.658 - 14.25*0.658^2)j = 0.81i + (3.74 - 6.17)j = (0.81i - 2.43j) m/s

(c) Acceleration, a = dv(t) / dt = 0 + (5.68 - 2*14.25t)j = (5.68 - 28.5t)j

So, a(0.826) = (5.68 - 28.5*0.826)j = (5.68 - 23.54)j = -17.86j m/s^2