A girl of mass m_1 = 60.0 kilograms springs from a trampoline with an initial up

Question

A girl of mass m_1 = 60.0 kilograms springs from a trampoline with an initial upward velocity of v_i = 8.00 meters per second. At height h = 2.00 meters above the trampoline, the girl grabs a box of mass m_2 = 15.0 kilograms. For this problem, use g = 9.80 meters per second per second for the magnitude of the acceleration due to gravity. What is the speed v_before of the girl immediately before she grabs the box? v_before = m/s What is the speed v_after off the girl immediately after she grabs the box? v_after = m/s

Explanation / Answer

conservation of energy:

potential energy gain = loss in kinetic energy

=> mgh = [½mv²]

=> gh = ½[8² - v²]

=> v² = -2(9.80)2 + 64 = 103.2

=> v = 4.98 m/s (upwards) ...............................................................ans a)

conservation of momentum(mv):

momentum before impact = after

=> 4.98(60) = 75v'

=> v' = 3.98 m/s (upwards) .............................................................................ans b)

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