Question
a body of mass3kg and body of mass 2kg are dropped simultaneously from height of 14.9 m calculate their momenta and their kinetic energies when they are 5m above the ground
Answers
Henry
Given:
M1 = 3kg.
M2 = 2kg.
h = 14.9m.
d = 14.9 - 5 = 9.9m Traveled.
V^2 = Vo^2 ) 2g*d = 0 + 19.6*9.9 = 194,
V = 13.93 m/s = Velocity of each body.
Momentum1 = M1*V.
Momentum2 = M2*V.
KE1 = 0.5M1*V^2.
KE2 = 0.5M2*V^2.
M1 = 3kg.
M2 = 2kg.
h = 14.9m.
d = 14.9 - 5 = 9.9m Traveled.
V^2 = Vo^2 ) 2g*d = 0 + 19.6*9.9 = 194,
V = 13.93 m/s = Velocity of each body.
Momentum1 = M1*V.
Momentum2 = M2*V.
KE1 = 0.5M1*V^2.
KE2 = 0.5M2*V^2.