Problem 3 A dump truck is sitting at the top of a hll, and its parking brake dis

Question

Problem 3 A dump truck is sitting at the top of a hll, and its parking brake disengages. It rolls down a hill and collides with an empty, stationary car. The car sticks to the grill of the truck and the two continue moving forward. The dump truck has mass (mt 4000kg) and is traveling at when it hits the car. The car has mass (mc 1500kg) (a)Why does the truck roll down the hill? Answer this very carefully using energy arguments. (b)What is the speed of the truck/car immediately after the collision? (c)What is the momentum of the truck/car in the frame of reference of a jogger approaching the collision at 2 (opposite the truck's initial velocity)? (d)How fast, and in which direction, would the jogger have to run for the momentum of the system to appear to him to be zero? (e)What is the momentum of the car and truck system prior to the collision in the center of mass frame? (f)Relative to an observer standing on the street, how much of the initial kinetic energy of the system is convertible? Does this answer depend on the frame of reference of the observer, why or why not? (g)Relative to an observer standing on the street, how much of the initial kinetic energy of the system must be conserved to conserve momentum? Does this answer depend on the frame of reference of the observer, why or why not?

Explanation / Answer

(a)

before barke fails , teh gravitationall force act on the truck is balcend by the frictional force , after barke fails , the gravitational force si greater than frictional force . due to unbalanced force , teh trcuk rolls

(b)

Apply conservation of momentum

mt ut + mc uc = ( mt + mc) v

4000(5) + 1500(0) = (5500) v

v= 3.63 m/s

(c)

p = ( mt + mc) v+ v' = (4000+ 1500) ( 3.63+2) = 31000 kgm/s

(d)

jogger have to move same velocity of the trcuk car system vj = 3.63 m/s

(e)

pi = mt vt = 4000(5) = 20000 kg m/s

(f)

del KE = KEf - kEi = 1/2 ( 4000+ 1500) ( 3.63^2 - 1/2 *( 4000) 5^2)= 13763.75 J

yes , the answer depend on the frame of refrence frame of observer

(g)

KEi = 1/2 * ( 4000+ 1500) ( 3.63^2 = 36236.47 J

yes,the answer depend on the frame of refrence frame of observer

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