Two 7.1 kg bodies, A and B, collide. The velocities before the collision are =(11+17) m/s and VB = : (−26î + 6.8ĵ) m/sAfter the collision, A = (-161 +8.8ĵ) m/s . What are (a) the x-component and (b) the y-component of the final velocity of B? (c)What is the change in the total kinetic energy (including sign)?(a) Number(b) NumberUnitsUnits(c) Number iUnits>

Answered: Two 7.1 kg bodies, A and B, collide.…

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter8: Momentum And Collisions

Section: Chapter Questions

Problem 1OQ

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Two skateboarders, with masses m1 = 75.0 kg and m2 = 65.0 kg, simultaneously leave the opposite sides of a frictionless half-pipe at height h = 4.00 m as shown in Figure P11.49. Assume the skateboarders undergo a completely elastic head-on collision on the horizontal segment of the half-pipe. Treating the skateboarders as particles and assuming they dont fall off their skateboards, what is the height reached by each skateboarder after the collision? FIGURE P11.49
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A rocket has total mass Mi = 360 kg, including Mfuel = 330 kg of fuel and oxidizer. In interstellar space, it starts from rest at the position x = 0, turns on its engine at time t = 0, and puts out exhaust with relative speed ve = 1 500 m/s at the constant rate k = 2.50 kg/s. The fuel will last for a burn time of Tb = Mfuel/k = 330 kg/(2.5 kg/s) = 132 s. (a) Show that during the burn the velocity of the rocket as a function of time is given by v(t)=veln(1ktMi) (b) Make a graph of the velocity of the rocket as a function of time for times running from 0 to 132 s. (c) Show that the acceleration of the rocket is a(t)=kveMikt (d) Graph the acceleration as a function of time. (c) Show that the position of the rocket is x(t)=ve(Mikt)ln(1ktMi)+vet (f) Graph the position during the burn as a function of time.

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