A 4000-kg dump truck is parked on a hill. The parking brake fails, and the truck

Question

A 4000-kg dump truck is parked on a hill. The parking brake fails, and the truck rolls down the hill. It then coasts briefly along a flat stretch of road at39 km/h before hitting a stationary 1000-kg car. The car sticks to the grill of the truck, and the two vehicles continue moving forward. Call the initial direction of motion of the truck the +x axis.

a)What is the x-component of the momentum of the combined unit in the reference frame of a jogger approaching the collision at 2.0 m/s and moving in the direction opposite the direction of the velocity of the moving unit?

b)Determine the x-component of the velocity with that the jogger must be running in order for the momentum of the combined unit to be zero in his reference frame?

Explanation / Answer

a) velocity of dump truck = 39 km/h = 39 x 1000m / 3600sec =10.83 m/s

velocity of dump truck wrt jogger = 10.83 - (-2) = 10.83 +2 = 12.83 m/s

velocity of car wrt jogger = 0 - (-2) = 2 m/s

momnetum (mv) =4000 x 12.83 + 1000 x 2 = 53333.33 kg m/s

b) velocity of combines unit = 10.83 x 4000 / (4000 + 1000) = 8.66 m/s

so jogger have to run with 8.66 m/s in the direction of velocity of combined unit.

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