1. Consider a model rocket that is launched in the vertical direction. Neglect a
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Question
1. Consider a model rocket that is launched in the vertical direction. Neglect air resistance and changes in its mass, m. Assume that the rocket engine delivers an impulse, Io which corresponds to a constant thrust. T0 applied over a time r . (a) From impulse considerations alone, show that the time to reach maximum altitude is ta =I0/mg (b) Determine the maximum altitude. (c) Show that if 1 mg, then the maximum altitude is (d) Estimate the altitude of a 2 oz. rocket propelled by a 5 Newton-second impulse. 2. A high-speed proton of electric charge e moves with constant speed v0 in a straight line past an electron of mass m and charge -e that is initially at rest. The electron is at a distance a from the path of the proton. (a) Assume that the proton passes so quickly that the electron does not have time to move appreciably from its initial position until the proton is far away. Show that the component of force in a direction perpendicular to the line along which the proton moves is given by (in mks units): e2a where t = 0 corresponds to when the proton passes closest to the electron. (b) Calculate the impulse delivered by this force. (c) Write the component of the force in a direction parallel to the proton velocity and show that the net impulse in that direction is zero. (d) Using these results, calculate the (approximate) final momentum and final kinetic energy of the electron. (e) Show that the condition for the original assumption in part (a) to be valid is e^2/4 Pi Eo aExplanation / Answer
Sorry we are allowed to solve one question at a time. So I am solving first one.
(a) We know that impulse = change in momentum
Since intial velocity is zero thefore initial momentum will be zero
So change in momentum = final - initial = mV where V is the final velocity
V = u+ at from first equation of motion
V = gta where accelereation is g
Now putting in the above equation
Inpulse = change in momentum
Io = m(gta)
ta = Io/mg
(b) For maximum altitude
S = ut + (1/2)at2
S = (1/2)g(Io/mg)2 = (Io/m)2(1/2g)
(c) If Thrust force is greater than mg
then we can negelect mg
therefore only the net acceleration is g
From second equation of motion
S = (1/2)gt2
(d) In this 2 oz is not clear
but as we know that impulse = change in momentum
therefore we can apply the same procedure as done in a part.
5 = mV
V = 5 /m
And from third equation of motion
V2 = u2 + 2aS
S = V2/2a = 25/2gm2
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