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

Question

(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.

The vector position of a 3.05 g particle moving in the xy plane varies in time according to jt. (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. It^2 ? 6i ? 2r2 = 3r with arrow is in centimeters. At the same time, the vector position of a 5.55 g particle varies as jt^2 where t is in seconds and j)t + 2i + 3r1 = (3

Explanation / Answer

[a]

For particle 1 : r1 = (3i + 3j)t + 2jt2

At t=2.60 s

r1 = (3i + 3j)*2.60 + 2j*2.602 =  [7.80i+21.32 j] cm

For particle 2 : r2 = 3i + 2it2 + 6jt

At t=2.60 s

r2 = 3i+2i*2.602+ 6j*2.60=  [16.52i+15.6j] cm

vector position of centre of mass= [m1r1+m2r2] / [ m1+m2] = 3.05 [7.80i+21.32 j] +5.55[16.52i+15.6j] / 8.6

= [23.79i+65.03j+91.69i+86.58 j] / 8.6= [115.48i+151.61j] / 8.6

= [13.43i+17.63 j] cm

[b]

p1= m1v1= 3.05 g *[3i+13.4 j] cm/s= [9.15i +40.87j] gcm/s

p2= m2v2= 5.55 g *[10.4i+6 j] cm/s= [57.72i +33.3j] gcm/s

[c]

For particle 1 : r1 = (3i + 3j)t + 2jt2
Differentiate the equation once with respect to time, we get :
v1 =d r1 /dt= (3i + 3j)+ 4jt

At t=2.60 s

v1 = (3i + 3j)+ 4j*2.60= [3i+13.4 j] cm/s

v1=[3i+13.4 j] cm/s

For particle 2 : r2 = 3i + 2it2 + 6jt
Differentiate the equation once with respect to time, we get :
v2 =d r2/dt= 4it+ 6j

At t=2.60 s

v2 = 4i*2.60+6j

v2= [10.4i+6 j] cm/s

velocity of centre of mass= [m1v1+m2v2] / [ m1+m2]

=[9.15i +40.87j+57.72i +33.3j] / 8.6= [66.87 i+74.17 j] / 8.6

= [7.78i+8.62 j] cm/s

[d] For particle 1 : r1 = (3i + 3j)t + 2jt2
Differentiate the equation twice with respect to time, we get :
a1 =d2r1 /dt2= 4j

At t=2.60 s; a1 = 4j

For particle 2 : r2 = 3i + 2it2 + 6jt

Differentiate the equation twice with respect to time, we get :
a2 =d2r2 /dt2= 4i

At t=2.60 s; a2 = 4i

acceleration of centre of mass= [m1a1+m2a2] / [ m1+m2]

= [3.05*4j+5.55*4i] / 8.6

= [2.58 i+ 1.42 j] cm/s2

[e]

Fnet =m1a1+m2a2 = [3.05*4j+5.55*4i] =[22.2i+12.2j] *10-5 N

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.