a) In outer space, far from other objects, block 1 of mass 39kg is at position (

Question

a) In outer space, far from other objects, block 1 of mass 39kg is at position ( 4, 11, 0) m, and block 2 of mass 1160kg is at position is the (vector) gravitational force acting on block 2 due to block 1? It helps to make a sketch of the situation. 20, 11, 0) m. What grav- Submit Answer Tries 0/5 At 4.3 seconds after noon both blocks were at rest at the positions given above. At 4.475 seconds after noon, what is the (vector) momentum of block 2? P2 Submit Answer Tries 0/5 At 4.475 seconds after noon, what is the (vector) momentum of block 1? Pi

Explanation / Answer

a) given

m1 = 39 kg, (4,11,0)

m2 = 1160 kg (20,11,0)

force on 2 due to 1 = F21 = Gm1*m2*r/|r|^3

r = (4 - 20)i + (11 - 11)j + (0 - 0)k = -16i

F21 = -G*39*1160 i/16^2 = -117.8714062 *10^-10 i N ( along -ve x axis)

b) at t = 0, both are at rest

ro = 16 m

at t = 0.175 s

at time t

distance between them = r

also

m1*r" = Gm1m2/r^2

r" = Gm2/r^2 = d^r/dt^2 = dvr/dt * dr/dr = vr*dvr/dr

Gm2*dr/r^2 = vr*dvr

integrating from r = ro to r = r, vr = 0 to vr

vr is relative speed of approach of the two objects

Gm2*(1/ro - 1/r) = vr^2/2

vr = sqrt(2Gm2(1/ro - 1/r)) = dr/dt

integrating again, repl;acing masses

sqrt(2Gm1)*t = -sqrt(a)[a*ln([sqrt((r-a)/r) + 1]/[sqrt((r-a)/r) - 1]) + 2r*sqrt((r-a)/r)]/2

at t= 0.175s

[16*ln([sqrt((r-16)/r) + 1]/[sqrt((r-16)/r) - 1]) + 2r*sqrt((r-16)/r)] = -6.3*10^-6

fropm graphical methods we see

r = 7.1*10^-4 m

hence

vr = sqrt(2Gm1(1/16 - 1/r)) = 0.0027068949871213 m/s

momentum = 1160*vr = 3.139998185 kg m/s

directino of momentum from 2 to 1 = alopng +i direction

hence

momentum =- 3.1399 i kg m/s

c) block 1

momentum = 3.1399 i kgm/s

because p1 + p2 = 0 ( fromconservation of moemntum)

and initialmomentuim of the system is 0

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