A 1,020lb space vehicle traveling with velocity Vo=(1570ft/s)k passes through th

Question

A 1,020lb space vehicle traveling with velocity Vo=(1570ft/s)k passes through the origin O. Explosive charges then separate the vehicle into three parts A, B, and C, with masses of 170lb, 340lb, and 510lb, respectively. Knowing that shortly thereafter the positions of the three parts are, respectively, A(250,250,2250), B(600,1300,3200), C(-475,-950,1900), where the coordinates are expressed in meters, that the velocity of B is Vb=(500ft/s)i + (1100ft/s)j + (2100ft/s)k, and that the x component of the velocity of C is -390ft/s, determine the velocity of part A.

Explanation / Answer

Conversion: 1 m = 3.28 ft

Momentum balance in x-direction: 1020*0 = 170*Vax + 340*500 + 510*(-390)

So, Vax = 170

Momentum balance in y-direction: 1020*0 = 170*Vay + 340*1100 + 510*Vcy

Momentum balance in z-direction: 1020*1570 = 170*Vaz + 340*2100 + 510*Vcz

Using distance = velocity*time for particle A we get, 250*3.28 = Vax*t = 170*t

t = 4.82 s

Similarly, 250*3.28 = Vay*t = Vay*4.82

Vay = 170.1 ft/s

Similarly, 2250*3.28 = Vaz*t

Vaz = 1531.12 ft/s

Va = (Vax^2 + Vay^2 + Vaz^2) = 1550 ft/s

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