- #1
jmlibunao
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Homework Statement
Two particles of masses 8 kg and 5 kg are connected to the two ends of a light inextensible string which passes over a fixed smooth pulley. Initially each of the two particles are held at a position which is 5 m above a horizontal ground. The objects are then released from rest. Assuming that the particles never reach the pulley, and also that the particles do not rebound when they strike the ground,
#1 Find the tension in the string and the acceleraion of each particle
#2 Find the distance above the ground of the point H, the highest point reached by the 5 kg mass;
#3 Find the speed with which the system is jerked into motion and the impulse experienced by the string during the jerk.
Homework Equations
g = 10 m/s^2
F1 = m1a = m1g - T
F2 = m2a = T - m2g
The Attempt at a Solution
The acceleration of both particles would be the same.
I can easily solve for the acceleration, a, and tension T by:
(8 kg)(a) = (8 kg)(10 m/s^2) - T -- (1)
(5 kg)(a) = T - (5 kg)(10 m/s^2) -- (2)
I'd let both (1) and (2) be equal to T and then solve for a, which I got to be 2.3. I would then plug this in either (1) or (2) and then solve for T, which I got to b 61.54
My question lies in #2 and #3.
For #2, wouldn't the highest point the 5 kg mass would reach be 10 m (5 m + 5 m)?
For #3, can I use one of the kinematics equation to solve for a velocity v? I'm considering the string itself to be massless, so the total mass of the system would be 8 kg + 5 kg = 13 kg. Wouldn't I be able to get the impulse by multiplying 13 kg by g = 10 m/s^2 and my computed v (assuming that I do compute for a v)?