07prob_11

# 07prob_11 - Assigment#7 Page 1 of 4 UC Davis Physics 9A-C...

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Unformatted text preview: Assigment #7, 5/13/11 Page 1 of 4 UC Davis Physics 9A-C Assignment #7 Cole [8.7]    Δ p Δ t = 0.045 kg ( ) 25 m ( ) 2 × 10 − 3 s  =  563 N. The weight of the ball is less than half a Newton, so the  weight is not significant while the ball and club are in contact. [8.20]  a) This process happens in the center of mass frame where  Í p =  0, both before and after  the collision.   m A ′ v A + m B ′ v B = .  õ     ′ v A = − m B ′ v B m A = − 3  kg 1  kg 1.2  m/s ( )  =  -3.6 m/s  b)  This is an inelastic "explosion", but it uses potential energy stored in the spring so we can  conserve total energy,  Eø = E f .  The potential energy of the spring appears as kinetic energy of  the masses:   U = K A   + K B   =   (1/2)(1.00 kg)(3.60 m/s) 2  + (1/2)(3.00 kg)(1.200 m/s) 2  = 8.64 J . [8.27]  As the rain hits the car, the water experiences an impulse in the direction of travel of the  car which speeds the water up so it travels horizontally with the car, and the car experiences and  equal but opposite impulse which slows the car down .  Conserving momentum Mvø  =  (m + M) v  õ v = 2.4 × 10 4 kg 4 m/s ( ) 2.7 × 10 4 kg  =  3.56 m/s.   [8.35]  a)  The small car places a force  F  on the large car, and by Newton's 3rd law, the large car  must place an equal but opposite force on the small car.   F = Δ p t  õ  They experience equal but  opposite momentum changes.  õ   m s Δ v s = m L Δ v L   õ  Δ v s = m L m s Δ v L = 3000 1200 Δ v L = 2.5 Δ v L b)  The acceleration is proportional to the velocity change; hence, the small car has 2.5 times the  acceleration of the large car.  The forces experienced by the small car passengers are 2.5 times  that of the large car.  This difference can be mitigated by building crumple zones into cars so the  any passengers wearing seat belts stop more slowly.  However, note that massive cars have a big  safety advantage just by virtue of their large mass. [8.37]  The initial momentum of the car must be the  x-component of the final momentum  as the truck had no initial  x-component of momentum, so   Í p x   = m car v car   =  (m car  + m truck )v  sin 24° v car = m car + m truck ( ) m car v sin24 ° = 2850 950 16 m/s ( ) sin24 °  = 19.5 m/s.  For the  y- direction:  Í p y...
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## This note was uploaded on 09/28/2011 for the course PHY 9A taught by Professor Svoboda during the Spring '08 term at UC Davis.

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07prob_11 - Assigment#7 Page 1 of 4 UC Davis Physics 9A-C...

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