2) what is the y position of the center of mass ? 3) what is the speed of the ce

Question

2) what is the y position of the center of mass ? 3) what is the speed of the center of mass ? ttps//www.tlipitphysics.com/Course/ViewProbl Deadline: 100% until Saturday, February 27 at 11:59 PM System of Particles System of Partices 2 3 Four particles are in a 2-D plane with masses, x- and y positions, and x- and y- velocities as given in the table below: 1 8.9 kg 2.4 m -4.5 m 2.9 m/s 4 m/S 8.9 kg 5.4 m 2.5 m 3.9 m/s 3 m/s 1) V/hat is the x position of the center of mass? 1.710 m Submit Your submissions: 1,710 X Computed value: 1.710 Feedback: ter of n

Explanation / Answer

given values are

m1=8.9 kg , x1=-2.4 m , y1=-4.5 m , Vx1=2.9 m/s , Vy1 = -4 m/s

m2 = 7.6 kg . x2 = -3.5 m , y2= 3.6 m , Vx2 = -5 m/s , Vy2 = 5.1 m/s

m3=8.2 kg , x3= 4.6 m , y3=-5.4 m, Vx3= -6.1 m/s .Vy3= 1.9 m/s

m4=8.9kg , x4=5.4 m, y4 = 2.5 m, Vx4= 3.9 m/s , Vy4 = -3 m/s

1. x - position of Centre of mass =( m1* x1 + m2*x2 + m3*x3 + m4*x4 ) /( m1 +m2+m3+m4)

x - position of Centre of mass = ((8.9*(-2.4)) +(7.6*(-3.5))+(8.2*4.6)+(8.9*5.4))/(8.9+7.6+8.2+8.9)

x - position of Centre of mass = 37.82/33.6 = 1.12 m

2. y - position of Centre of mass =( m1* y1 + m2*y2 + m3*y3 + m4*y4 ) /( m1 +m2+m3+m4)

y - position of Centre of mass = ((8.9*(-4.5)) +(7.6*(3.6))+(8.2*-5.4)+(8.9*2.5))/(8.9+7.6+8.2+8.9)

y - position of Centre of mass = -1.03 m

3. Vx of the centre of mass = ( m1* Vx1 + m2*Vx2 + m3*Vx3 + m4*Vx4 ) /( m1 +m2+m3+m4)

Vx of the centre of mass = ((8.9*(2.9)) +(7.6*(-5))+(8.2*-6.1)+(8.9*3.9))/(8.9+7.6+8.2+8.9)

Vx of the centre of mass =   -0.81 m/s

Vy of the centre of mass = ( m1* Vy1 + m2*Vy2 + m3*Vy3 + m4*Vy4 ) /( m1 +m2+m3+m4)

Vy of the centre of mass = ((8.9*(-4)) +(7.6*(5.1))+(8.2*1.9)+(8.9*-3))/(8.9+7.6+8.2+8.9)

Vy of the centre of mass = -0.23 m/s

therefore speed = sqrt (Vy^2 + Vx^2)

speed = sqrt (0.054 + 0.65) = sqrt(0.71)

speed = 0.84 m/s

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