The vector position of a 360 g particle moving in the xy pla

The vector position of a 3.60 g particle moving in the xy plane varies in time according to r with arrow1 = 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 arrow2 = 3i 2it2 6jt.

(a) Determine the vector position of the center of mass at t = 2.60.

(b) Determine the linear momentum of the system at t = 2.60.

(c) Determine the velocity of the center of mass at t = 2.60.

(d) Determine the acceleration of the center of mass at t = 2.60.

(e) Determine the net force exerted on the two-particle system at t = 2.60.

Solution

Given :

Arrow 1 …..mass = 3.60 g

   r ₁ = 3i + 3j t + 2jt²

Arrow 2…… mass = 5.45 g

r₂ = 3i 2it² 6jt

Now ,

a)   the vector position of the center of mass at t = 2.60 is,

center of mass = (m₁r₁ +m₂r₂)/(m₁ + m₂)…………..equation 1

Now, m₁r₁ =(3.60)*( 3i + 3j (2.60) + 2j(2.60) ² at t = 2.60 s

                 = (3.60)*( 3i + 7.8j + 13.52j )

                 =10.8i + 28.8j + 48.6j

                 =10.8i + 77.4j…………………………..equation 2

         m₂r₂ = (5.45)*( 3i 2i(2.60) ² 6j(2.60)

                = (5.45)*( 3i – 13.52i – 15.6j)

                =(5.45)*( – 10.52i – 15.6j)

                = - 57.3i – 85.02j……………………….equation 3

Add equation 2 and equation 3, we get

(m₁r₁ +m₂r₂)= (10.8i + 77.4j) + (- 57.3i – 85.02j)

                   = - 46.5i – 7.8j………………………….Equation 4

                 And

            (m₁ + m₂) = (3.60 +5.45) = 9.05 g…………….equation 5

Divide equation 4 by equation 5,

center of mass = (m₁r₁ +m₂r₂)/(m₁ + m₂)

                        = (- 46.5i – 7.8j)/ 9.05

                         = - 5.13i – 0.86j

linear momentum of the system = P1 + P2

                                                   = m₁v₁ + m₂v₂

Where,    v₁ = velocity of first body

                v₁ = dr₁/dt = 3j +4jt

at t = 2.60 s,

v₁ = 3j +4j(2.60)

       =3j +10.4j

        =13.4j……………. …………………..equation 6

So linear momentum of first body is,

P1 = m₁v₁ = (3.60)*( 13.4j) = 48.24j ………… equation 7

                Now,

v₂ = velocity of second body

v₂ = dr₂ /dt = - 4it – 6j

at t = 2.60 s,

v₂ = dr₂ /dt = - 4i(2.60) – 6j = - 10.4i – 6j

Therefore,

P2 = m₂v₂ = (5.45)*( - 10.4i – 6j) = -56.68i – 32.7j………. equation 8

Now add equation 7 and equation 8, we get

linear momentum of the system = P1 + P2

= 48.24j + (-56.68i – 32.7j) = - 56.68i + 15.54j

c)   Determine the velocity of the center of mass at t = 2.60.

       center of mass = (m₁v₁ +m₂v₂)/(m₁ + m₂)

                                =48.24j + (-56.68i – 32.7j) / 9.05

                              =( - 56.68i + 15.54j)/9.05

                               = - 6.26i + 1.71 j

Acclerations can be calculate by using the formula,

A1 = dv₁ /dt = 7j,

And A2 = dv₂ /dt = - 4i – 6j

center of mass = (m₁A₁ +m₂A₂)/(m₁ + m₂)

= (25.2j + (- 21.8i – 32.7j)) / 9.05

= (- 21.8i - 7.5j ) /9.05

= 2.40 - 0.82 j

Now from above calculation, we can observe that all equations are independent on time. So accleration is same at every instant. That means it is uniform.

e)   Determine the net force exerted on the two-particle system at t = 2.60

F = mass *accleration = (m₁A₁ +m₂A₂) = (25.2j + (- 21.8i – 32.7j))

F = (- 21.8i - 7.5j )

It is also independent on time. So this force is also uniform.