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Unformatted text preview: Ph1a Solutions 7 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. 7.1 QP8 (5 points) 7.1.a (0.5 points) The position of the ball at time t is given by x b ( t ) = v t . The center of mass is therefore x cm ( t ) = Σ m i x i Σ m i = mv t + Mx 1 m + M 7.1.b (1 point) The momentum before the collision is vector p i = mv ˆ x . After the collision vector p f = ( m + M ) v f ˆ x . Conserving momentum and solving: v f = m m + M v . 7.1.c (1.5 points) Any kinetic energy lost in the collision will be stored in the spring as potential energy. Initially, E i = 1 2 mv 2 . After the collision, E f = 1 2 ( m + M ) v 2 f + U spring . Conservation of energy yields U spring = 1 2 mv 2- 1 2 m m + M mv 2 = 1 2 mv 2 parenleftbigg M m + M parenrightbigg = parenleftbigg M m + M parenrightbigg E i 7.1.d (0.5 points) From part c: U spring = 1 2 mv 2 parenleftbigg M m + M parenrightbigg 1 2 k (Δ l ) 2 = 1 2 mv 2 parenleftbigg M m + M parenrightbigg Δ l = radicalBigg v 2 k mM m + M 7.1.e (1.5 point) The center of mass is unaffected by the motion of the spring and will continue moving to the right with velocity v f = m m + M v . The two masses will oscillate. The graphs are given in Fig. ?? . The full solution follows below, although the problem only asks for a qualitative solution....
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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