Question

The vector position of a 3.80 g particle moving in the xy plane varies in time according to
r (with arrow)1 = (3i + 3j)t + 2jt2 where t is in seconds and r with arrow is in centimeters. At the same time, the vector position of a 5.45 g particle varies as r (with arrow)2 = 3i − 2it2 − 6jt.

(a) Determine the vector position of the center of mass at t = 2.90.

(b) Determine the linear momentum of the system at t = 2.90.

(c) Determine the velocity of the center of mass at t = 2.90.

(d) Determine the acceleration of the center of mass at t = 2.90.

(e) Determine the net force exerted on the two-particle system at t = 2.90.

Answer 1

at t = 2.9 sec

r1 = (3i+3j)*2.9 + (2*2.9^2)j

r1 = 8.7i + 25.52 j

r2 = 3i - (2*2.9^2)*i -6*2.9*j

r2 = -13.82 i - 17.4 j

Xcm = (m1*x1)+(m2*x2)/(m1+m2)

Xcm = ((3.8*8.7)-(5.45*13.82))/(3.8+5.45) = -4.56 cm

Yxm = ((3.8*25.52)-(5.45*17.4))/(3.8+5.45) =0.232 cm

vector position of center of mass is r = -4.56 i + 0.232 j

b) v1 = dr1/dt = (3i+3j)+(4jt) = 3i + 3j + (4*2.9*j) = 3i + 14.6 j

v2 = -4it -6j

v2 = -(4*2.9*i) -6j

v2 = -11.6 i - 6j

Vx-CM = ((3.8*3)-(5.45*11.6))/(3.8+5.45) = -5.6 cm/sec

Vy-CM = ((3.8*14.6)-(5.45*6))/(3.8+5.45) = 2.46 cm/sec

v = sqrt(2.46^2+5.6^2) = 6.12 cm/sec

linear momentum is p = m*v = (3.8+5.45)*10^-3*6.12*10^-2 = 5.661*10^-4 kg-m/sec

C) V_CM = 6.12 cm/sec = 0.0612 m/sec

D) ax-CM = dvx/dt = -4i

ay-CM = 4j

a = -4i + 4j

a = sqrt(4^2+4^2) = 5.65 cm/s^2

e) F = m*a = (3.8+5.45)*10^-3*5.65*10^-4 = 5.22*10^-6 N