Problem#1
Consider as a system the Sun with the Earth in a circular orbit around it. Find the magnitude of the change in the velocity of the Sun relative to the center of mass of the system over a period of 6 months. Neglect the influence of other celestial objects. You may obtain the necessary astronomical data from the endpapers of the book.
Answer:
The orbital speed of the Earth is
vE = 2πr/T
vE = 2π(1.496 x 1011 m)/(3.156 x 107s) = 2.98 x 104 m/s
In six months the Earth reverses its direction, to undergo momentum change
mE│∆vE│ = 2mEvE = 2(5.98 x 1024kg)(2.98 x 104 m/s) = 3.56 x 1026 kg.m/s
Relative to the center of mass, the sun always has momentum of the same magnitude in the
opposite direction. Its 6-month momentum change is the same size,
mS│∆vS│ = 3.56 x 1026 kg.m/s, then
│∆vS│ = (3.56 x 1026 kg.m/s)/(1.991 x 1030kg) = 0.179 m/s
Problem#2
Review problem. There are (one can say) three coequal theories of motion: Newton’s second law, stating that the total force on an object causes its acceleration; the work–kinetic energy theorem, stating that the total work on an object causes its change in kinetic energy; and the impulse–momentum theorem, stating that the total impulse on an object causes its change in momentum. In this problem, you compare predictions of the three theories in one particular case. A 3.00-kg object has velocity 7.00j m/s. Then, a total force 12.0i N acts on the object for 5.00 s. (a) Calculate the
object’s final velocity, using the impulse–momentum theorem. (b) Calculate its acceleration from a = (vf - vi)/∆t. (c) Calculate its acceleration from a = ∑F/m. (d) Find the object’s vector displacement from ∆r = vit + ½ at2 (e) Find the work done on the object from W = F.∆r, (f) Find the final kinetic energy from ½ mvf2 = ½ mvf.vf, (g) Find the final kinetic energy from ½ mvf2 + W.
Answer:
(a) we get
Ft = ∆p = m(vf – vi)
(12.0Ni)(5.00s) = (3.00kg)(vf – 7.00j m/s)
60.0i kg.m/s = 3.00kgvf – 21.0j kg.m/s
vf = (20.0i + 7.00j)m/s
(b) a = (vf - vi)/∆t
a = [(20.0i + 7.00j)m/s – 7.00j m/s]/(5.00s) = 4.00i m/s2
(c) a = ∑F/m = 12.0iN/(3.00kg) = 4.00i m/s2
(d) ∆r = vit + ½ at2
∆r = (7.00j m/s)(5.00s) + ½ (4.00i m/s2)(5.00s)2
∆r = (50.0i + 35.0j) m
(e) W = F.∆r = (12.0iN).(50.0i + 35.0j)m = 600J
(f) ½ mvf2 = ½ mvf.vf = ½ (3.00kg)(20.0i + 7.00j)(20.0i + 7.00j)m2/s2
½ mvf2 = 674J
(g) ½ mvi2 + W = ½ (3.00kg)(7.00m/s)2 + 600J = 674 J
Problem#3
A rocket has total mass Mi = 360 kg, including 330 kg of fuel and oxidizer. In interstellar space it starts from rest. Its engine is turned on at time t = 0, and it puts out exhaust with relative speed ve = 1500 m/s at the constant rate 2.50 kg/s. The burn lasts until the fuel runs out, at time 330 kg/(2.5 kg/s) = 132 s. Set up and carry out a computer analysis of the motion according to Euler’s method. Find (a) the final velocity of the rocket and (b) the distance it travels during the burn.
Answer:
(a) the final velocity of the rocket is
v(t) = –veln[1 – kt/Mi]
vf = –(1500m/s)ln[1 – (2.50 kg/s)(132s)/(360 kg)]
vf = 3.73 km/s
(b) the distance it travels during the burn is
x(t) = ve(Mi/k – t) ln(1 – kt/Mi) + vet
xf = (1500 m/s)[360 kg/(2.50kg/s) – 132s] ln[1 – (2.50kg/s)(132 s)/(360kg)] + (1500 m/s)(132s)
xf = 153 km
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