Question

In outer space a rock of mass 12 kg is acted on by a constant net force 21, −11, 44 N during a 4 s time interval. At the end of this time interval the rock has a velocity of 107, 81, 104 m/s. What was the rock's velocity at the beginning of the time interval?

Answer 1

Answer:

(100,84.67,89.33) m/s

Explanation:

From Newton's second law,

Impulse = Final momentum- initial momentum.

I = Mf-Mi.............. Equation 1

make Mi the subject of the equation

Mi = Mf-I............ Equation 2

Where I = Impulse. Mf = final momentum, Mi = initial momentum.

But

I = Ft.................... Equation 3

Where F = Force, T = time

Given: F = (21,-11,44) N, t = 4 s.

I = 4(21,-11,44)

I = (84,-44,176) N.s

Also,

Mf = mVf................ Equation 4

Where m = mass, Vf = Final velocity

Given: m = 12 g, Vf = (107,81,104) m/s.

Substitute into equation 4

Mf = 12(107,81,104)

Mf = (1284,972,1248) kgm/s.

Substitute the value of I and Mf into equation 2

Mi = (1284,972,1248)-(84,-44,176)

Mi = (1200,1016,1072) kgm/s

Also,

Mi = mVi

Where Vi = Initial velocity.

make Vi The subject of the equation

Vi = Mi/m..........................  Equation 5

Given: m = 12 kg, Mi = (1200,1016,1072) kgm/s

Substitute into equation 5

Vi = (1200,1016,1072)/12

Vi = (100,84.67,89.33) m/s