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 at v_1 = 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 a the speed v_before of the girl immediately before she grabs the box? Express your answer numerically In meters per second. v_before = m/s What is the speed v_after of the girl immediately after she grabs the box? Express your answer numerically in meters per second v_after =

Explanation / Answer

A)
use conservation of energy ,
initial kinetic energy = final kinetic energy + final potential energy
0.5*m*vi^2 = m*g*h + 0.5*m*vbefore^2
0.5*vi^2 = g*h + 0.5*vbefore^2
0.5*8^2 = 9.8*2 + 0.5*vbefore^2
vbefore = 4.98 m/s

B)
use conservation of momentum,
mbefore*vbefore = m after * vafter
60*4.98 = (60+15)*vafter
vafter = 3.98 m/s

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