This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Ph1a Solutions 6 Chefung Chan ([email protected]), Fall 2007 Each homework problem is worth 5 points. Please disregard the point values listed on the problem itself. Use these instead. 6.1 QP7 (5 points) 6.1.a (1 point) The total kinetic energy just before the collision is equal to the total potential energy before the masses are released. K = U m + U M = ( m + M ) gh 6.1.b (2 points) For inelastic collisions, momentum is conserved and kinetic energy is not conserved. Define v = √ 2 gh and the velocity after the collision as v f . Conservation of momentum gives mv + M ( v ) = ( m + M ) v f v f = m M m + M v The resulting kinetic energy ( K f ) is K f = 1 2 ( m + M ) v 2 f = ( m M ) 2 ( m + M ) gh. Defining the maximum height after the collision as h ′ , conservation of energy (after the collision) gives ( m + M ) gh ′ = ( m M ) 2 ( m + M ) gh h ′ = parenleftbigg m M m + M parenrightbigg 2 h 6.1.c (2 points) Elastic collisions conserve both momentum and kinetic energy. Defining the velocities of the masses after the collision as v m and v M gives final momentum and kinetic energy of p f = mv m + Mv M K f = 1 2 mv 2 m + 1 2 Mv 2 M The conservation laws give ( m M ) v = mv m + Mv M 1 2 ( m + M ) v 2 = 1 2 mv 2 m + 1 2 Mv 2 M Solving these equations for the velocity of the small mass v m = m 3 M m + M v K m = 1 2 mv 2 m = parenleftbigg...
View
Full
Document
This note was uploaded on 06/05/2008 for the course PH 1a taught by Professor Goodstein during the Fall '07 term at Caltech.
 Fall '07
 Goodstein
 Energy, Kinetic Energy, Work

Click to edit the document details