physics hw - Display in a New Window Explore An object of...

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Display in a New Window Explore An object of mass m 1 = 10 kg and velocity v 1 = 1 m/s crashes into another object of mass m 2 = 20 kg and velocity v 2 = -17 m/s. The two particles stick together as a result of the collision. Determine the velocity v f of the objects after collision, and determine the change in total mechanical energy. Conceptualize The two particles with the masses given approach each other and then stick together, continuing their motion as if they were now one particle whose mass is the sum of the previous two masses. Categorize The problem concerns a collision that occurs in one dimension. The system consists of the two objects and the only forces affecting the motion are the internal forces between them during collision.
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(A) The velocity after collision. Analyze Because no external force acts, the collision does not change the total momentum of the system of two particles. We set the total momentum before collision to the total momentum afterward: m 1 v 1 + m 2 v 2 = ( m 1 + m 2 ) v f The final velocity of the two objects joined together is given by the previous momentum conservation equation: (1) v f = m 1 v 1 + m 2 v 2 m 1 + m 2 (2) V f = ( 10 kg)( 1 m/s) + ( 20 kg)( -17 m/s) ( 10 kg) + ( 20 kg) = m/s (B) The change in total mechanical energy Analyze While the forces that the two objects exert on each other cannot change their total momentum, they can change the total kinetic energy in an inelastic collision such as the one being considered. All of the mechanical energy in the problem is kinetic energy. Change in kinetic energy is then: Δ K = ½( m 1 + m 2 ) v f 2 - (½ m 1 v 1 2 + ½ m 2 v 2 2 ) At this point, we can either substitute the numerical values to evaluate Δ K , or we could substitute in Equation (2) for v f into the expression for Δ K and simplify the expression before doing the calculation. The first procedure is easier if only a numerical answer is needed, while the second is more useful in exploring how the loss in kinetic energy depends on the velocities and masses of the colliding objects. Directly substituting in numerical values gives: Δ K = ½( 10 kg + 20 kg)( -11.00 m/s) 2 - (½( 10 kg)( 1 m/s) 2 + ½( 20 kg)( -17 m/s) 2 ) = J.
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Following the second procedure instead leads to: Δ K = - ½ ( m 1 m 2 m 1 + m 2 ) ( v 1 - v 2 ) 2 This could be used to calculate the same numerical answer, but it also shows various features that you might examine in the simulation. For example notice that the change in kinetic energy
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This note was uploaded on 02/16/2010 for the course PHYS 1500 taught by Professor Loch during the Spring '08 term at Auburn University.

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physics hw - Display in a New Window Explore An object of...

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