Hmwk_07_Solutions

Hmwk_07_Solutions - Physics 221 Fall 2008 Homework #7...

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1 Physics 221 Fall 2008 Homework #7 Solutions Ch. 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 - z plane. The x -component of the force F x that 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 F x = 2.00 N. (a) What are the initial momentum and kinetic energy of the particle before the collision? The initial momentum is p 1 x = mv 1 x = (0.0100 kg)( ± 3.00 m/s) = ± (0.0300 kg m/s) . The initial kinetic energy is K 1 = (1/2) mv 1 x 2 = (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? J x = F x dt ± = F x ² t = (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? J x = p 2 x p 1 x . Therefore, p 2 x = p 1 x + J x = ± (0.0300 kg m/s) + 0.0600 kg m/s = 0.0300 kg m/s . The final velocity is v 2 x = p 2 x / m = (0.0300 kg m/s)/(0.0100 kg) = 3.00 m/s . The final kinetic energy is K 2 = (1/2) mv 2 x 2 = (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 . F x (N) time (s) 0 1 2 0 0.01 0.02 0.03 0.04
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2 7.2 Consider a system of N particles which exert forces on each other. These are called internal forces of the system. In addition, consider that external forces are exerted on the particles, where “external forces” are forces exerted on the N particles by particles that are outside of the defined system. Explain why the sum of all the forces on all N particles of the system only consists of the sum of all the external forces, i.e. , explain why the sum of all the internal forces add to zero. An internal force is one between a pair of particles within the system. From Newton’s 3 rd law, the two forces between each pair of particles exactly add to zero. Thus the sum of all internal forces in a system add to zero.
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This note was uploaded on 01/28/2010 for the course PHYSICS 221 taught by Professor Johnson during the Fall '06 term at Iowa State.

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Hmwk_07_Solutions - Physics 221 Fall 2008 Homework #7...

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