1 Physics 221 Fall 2008 Homework #7 SolutionsCh. 8 Due Tues, Oct 14, 2008 7.1 A 0.0100 kg particle is moving horizontally towards the negative x-axis at a speed of 3.00 m/s and collides with a vertical stationary wall in the y-zplane. The x-component of the force Fxthat the wall exerts on the particle during the collision is plotted in the figure. The particle is in contact with the wall during the time interval between 0.0100 s and 0.0400 s, during which Fx= 2.00 N. (a) What are the initial momentum and kinetic energy of the particle before the collision? The initial momentum is p1x= mv1x= (0.0100 kg)(±3.00 m/s) = ±(0.0300 kg m/s).The initial kinetic energy is K1= (1/2)mv1x2= (1/2)(0.0100 kg)(±3.00 m/s)2= 0.0450 J. (b) What is the impulse exerted by the wall on the particle? Jx=Fxdt±=Fx²t=(2.00 N)(0.0300 s)=0.0600 kg m/s. (c) What are the final momentum, velocity, and kinetic energy of the particle after colliding with the wall? Jx= p2x– p1x. Therefore, p2x= p1x+ Jx= ±(0.0300 kg m/s) + 0.0600 kg m/s = 0.0300 kg m/s . The final velocity is v2x= p2x/m= (0.0300 kg m/s)/(0.0100 kg) = 3.00 m/s . The final kinetic energy is K2= (1/2)mv2x2= (1/2)(0.0100 kg)(3.00 m/s)2= 0.0450 J. (d) Is the collision elastic or inelastic? Explain. From parts (a) and (c), the initial and final total kinetic energies are the same. Thus by definition, the collision of the particle with the wall is elastic. Fx(N)time (s)01200.010.020.030.04
has intentionally blurred sections.
Sign up to view the full version.