Rockets are propelled by the ejection of the products of the combustion of fuel. Consider a rocket of total mass m, to be travelling at speed v, in a region of space where gravitational forces are negligible. Suppose that the combustion products are ejected at a constant speed v, relative to the rocket. Show that a fucl 'burn' which reduces the total mass of the rocket to mą results in an increase in the speed of the rocket to vz, such that Uz - V1 = v, In " m2 Suppose that 2.1 x 10° kg of fuel are consumed during a 'burn' lasting 1.5 x 10 s. Given that there is a constant force on the rocket of 3.4 x 10' N during this 'burn', calculate v,. Hence calculate the increase in speed resulting from the 'burn' if m, is 2.8 x 10° kg. What is the initial vertical acceleration that can be imparted to this rocket when it is launched from Earth, if the initial mass is 2.8 × 10° kg?
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- Two manned satellites approaching one another at a relative speed of 0.100 m/s intend to dock. The first has a mass of 5.00 ✕ 103 kg, and the second a mass of 7.50 ✕ 103 kg. Assume that the positive direction is directed from the second satellite towards the first satellite. (a) Calculate the final velocity after docking, in the frame of reference in which the first satellite was originally at rest.m/s(b) What is the loss of kinetic energy in this inelastic collision?J(c) Repeat both parts, in the frame of reference in which the second satellite was originally at rest.final velocitym/sloss of kinetic energyJShow that for two inertial reference systems Sand S' that move with a speed U with respect to Si, does the following equality hold? 22 (√²-√²³) = 22 (√²² - √:²³) Where Ve and Vo are the initial and final velocities in the S system and V'f and Vo are the initial and final velocities in the sl system and m is the mass of the particle. Under what conditions is equality fulfilled?Suppose an object of mass m, initially travelling with a velocity v1, collides with an object of mass m initially at rest. Prove that the angle between the velocity vectors of the two objects of the two particles after the collision is 90 degrees.
- We have two spheres (m and M) that are separated by a small distance; m is to the left of M. A sphere of mass m (identical to one of the two initial spheres) is moving towards m at a speed V0. Show that when M is smaller than or equal to m, there will be 2 collisions and calculate the final speeds. Show that when M is larger than m, there will be three collisions and calculate the final velocities.A spacecraft cruising in space at a constant velocity of 2000 ft/s has a mass of 25,000 lbm. To slow down the spacecraft, a solid fuel rocket is fired, and the combustion gases leave the rocket at a constant rate of 150 lbm/s at a velocity of 5000 ft/s in the same direction as the spacecraft for a period of 5 s. Assuming the mass of the spacecraft remains constant, determine (a) the deceleration of the spacecraft during this 5-s period, (b) the change of velocity of the spacecraft during this time period, and (c) the thrust exerted on the spacecraft.Vahé and Nicolás, both with mass M, are playing catch with a football of mass m on the frictionless surface of a frozen pond. Since they're physics professors and not professional football players, they can only throw the football with a speed vo be only in terms of m, M, and vo- relative to themselves. All of your answers below should (а) Relative to a stationary observer on the surface of the pond, what is the speed v, with which the football is moving after it is Vahé throws the ball to his right, toward Nicolás. m M V |NM thrown? (b) Nicolás catches the ball. What is the speed v, with which Nicolás and the ball are moving after he catches the ball?
- 4. A rocket accelerates by burning its onboard fuel, so that its mass decreases with time.Suppose that the initial mass of the rocket at liftoff (including its fuel) is m, the fuel isconsumed at rate r, and the exhaust gases are ejected with constant velocity ν (relativeto the rocket). A model for the velocity of the rocket at time t is given by the equationv(t) = −ν ln(1 −Rt) −gt,where g is the acceleration due to gravity, R = r/m, and t is not too large.(a) Find an expression for the acceleration, a(t) = v′(t), of the rocket at any time t.(b) Find an expression for the position of the height, h(t), of the rocket at any timet. You can assume that at t = 0, the rocket hasn’t left the ground. Hint: re-member that ν,R,g are all constants and t is the independent variable.Write out your steps carefully!(c) Let g = 9.8 m/s2, R = 0.005 s−1. What does the velocity of the exhaust gases,ν, need to be for the rocket to reach a height of 6000 m one minute after liftoff?You can round to the…(b) A particle of unit mass moving in one dimension obeys the equation of motion x = (i) Show that i2 – e" is a constant of the motion. (ii) At t = 0 the particle is at x = 0 and at rest, i.e. x(0) = ¿(0) = 0. Show that the particle reaches x = +o at finite t > 0. Hint: use the result of part (i) to write the time taken to reach x = +0 as an integral. (iii) What is a suitable Lagrangian for this system?Two manned satellites approaching one another at a relative speed of 0.150M/S intend to dock. The first has a mass of 3.00×10^3 kg, the second a mass of 7.50 x 10^3 kg . Assume that the positive direction is directed from the second satellite towards the first satellite. (a) Calculate the final velocity after docking, in the frame of reference in which the first set a lot was originally at rest. m/s ? (b) what is the loss of kinetic energy in this inelastic collision? j ? (c) repeat both parts in the frame of reference in which the second satellite was originally at rest. Final velocity m/s ? loss of kinetic energy j ? explain why the change in velocity is different in the two frames, where areas the change in kinetic energy is the same in both. I used ^ to show exponents
- Consider a rocket traveling in a straight line subject to an external force Fext acting along the same line. a) Show that the equation of motion is mú = -mvext + Feat (1) b) Specialize to the case of a rocket taking off vertically (from rest) in a (constant) gravitational field g, so the equation of motion becomes múmvext mg (2) Assume that the rocket ejects mass at a constant rate, m = -k (where k is a positive constant), so that m = mg - kt Solve the equation for v as a function of t.7) Two manned satellites approaching one another, at a relative speed of 0.550 m/s, intending to dock. The first has a mass of 5.00 ✕ 103 kg, and the second a mass of 7.50 ✕ 103 kg. (a) Calculate the final velocity (after docking) in m/s by using the frame of reference in which the first satellite was originally at rest. (Assume the second satellite moves in the positive direction. Include the sign of the value in your answer.) _____m/s (b) What is the loss of kinetic energy (in J) in this inelastic collision? _____ J (c) Repeat both parts by using the frame of reference in which the second satellite was originally at rest. final velocity (m/s) _____ m/s loss of kinetic energy (J) _____ J Explain in detail why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both.7) Two manned satellites approaching one another, at a relative speed of 0.550 m/s, intending to dock. The first has a mass of 5.00 ✕ 103 kg, and the second a mass of 7.50 ✕ 103 kg. (a) Calculate the final velocity (after docking) in m/s by using the frame of reference in which the first satellite was originally at rest. (Assume the second satellite moves in the positive direction. Include the sign of the value in your answer.) _____m/s (b) What is the loss of kinetic energy (in J) in this inelastic collision? _____ J (c) Repeat both parts by using the frame of reference in which the second satellite was originally at rest. final velocity (m/s) _____ m/s loss of kinetic energy (J) _____ J Explain in detail why the change in velocity is different in the two frames, whereas the change in kinetic energy is the same in both.