Two objects (m1 = 5.50 kg and m2 = 2.70 kg) are connected by a light string pass

Question

Two objects (m1 = 5.50 kg and m2 = 2.70 kg) are connected by a light string passing over a light, frictionless pulley as in the figure below. The 5.50-kg object is released from rest at a point h = 4.00 m above the table.

Two objects (m1 = 5.50 kg and m2 = 2.70 kg) are connected by a light string passing over a light, frictionless pulley as in the figure below. The 5.50-kg object is released from rest at a point h = 4.00 m above the table. (a) Determine the speed of each object when the two pass each other. (b) Determine the speed of each object at the moment the 5.50-kg object hits the table. (c) How much higher does the 2.70-kg object travel after the 5.50-kg object hits the table?

Explanation / Answer

F=W1 - W2 (no inertia or weight of the string)
(m1+m2)a=(m1-m2)g

a=g(m1-m2)/(m1+m2)

a)They will pass each other with speed
V=at
when
h - distance passed by m1 on the way down
H-h - distance passed by m2 on the way up where H=4.00m

H-h=h or h=(1/2)H

h =2 m

a = g[(m1-m2)/(m1+m2)]

a = 3.35m/s^2
h=(1/2)at^2
t=sqrt(2h/a) or
t=sqrt(H/a)

t = 1.093 sec

V=at = 3.35*1.093

v = 3.662 m/s

b) Similarly to a accept t=sqrt(2H/a)
or use energy relationship

Pe1-Pe2=Ke1+Ke2
g(m1-m2)H=(1/2) (m1+m2)V^2
V=sqrt(2g(m1-m2)H/(m1+m2))

v = sqrt(2*9.8(5.5-2.7)*4/(5.5+2.7))

v = 5.174 m/s
c) Ke=Pe
(1/2) m2 V^2=m2 g y
y=V^2/(2 g )

y =26.77/2*9.8

y =1.37 m

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