I need help plz The vector position of a 3.20 g particle moving in the xy plane

Question

I need help plz

The vector position of a 3.20 g particle moving in the xy plane varies in time according to r1 = (3i + 3j)t + 2jt^2 where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.80 g particle varies as r2 = 3i - 2it^2 - 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.

Explanation / Answer

Given that

The mass of the first particle is (m1) =3.2*10-3kg

The position vecot is (r1) =(3i+3j)t+2jt2

The mass of the second particle is (m2) =5.80*10-3kg

The position vector of the second particle is(r2) =3i-2it2-6jt

The given positions of the first particle at time t =2.90s is r1 =(3i+3j)t+2jt2 =3*2.90i+3(2.90)j+2(2.90)2j=8.7i+25.52j

The position vector of the second particle is(r2) =3i-2it2-6jt at the time t =2.90s is r2 =3i-2it2-6jt =3i -2i(2.90)2 -6j(2.90) =-13.82i-17.4j

a)

The center of mass at t =2.90 is given by

rcm =m1r1+m2r2/(m1+m2) =(3.2*10-3kg)(8.7i+25.52j)+(5.80*10-3kg)(-13.82i-17.4j)/(3.2*10-3kg+5.80*10-3kg)=0.02784i+0.0816j-0.08015i-0.10092j/9*10-3=(-0.052i-0.0193j)m =(-5.21i-1.93j)cm

b)

The linear momentum of the system is given by

v1 =dr1/dt =(3i+3j)+4jt

at t =2.90s then v1 =(3i+3j)+4j(2.90) =3i+3j+11.6j=3i+14.6j

v2 =dr2/dt =-4it-6j

at t =2.90s then v2 =-4i(2.90)-6j =-11.6i-6j

Then the momentum is given by P =m1v1+m2v2 =(3.2)(3i+14.6j)+(5.80)(-11.6i-6j) =9.6i+46.72i-67.28i-34.8j=-57.68i+11.92j g.cm/s

c)

The velocity of center of mass is given by

Vcm =m1v1+m2v2/(m1+m2)

d)

The accelerataion of the first particle is

a1 =dv1/dt =4j

The acceleration of the second particle is

a2 =-4i

acm =m1a1+m2a2/(m1+m2)

similalry we have to find for vcm and acm

e)

The net force is given by Fnet =m1a1+m2a2 =(3.2)(4j)+(5.80)(-4i) =12.8j-23.2i

-23.2i+12.8j dynes

-232i+128j *10-6N

=-232i+128juN

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