The vector position of a 3.50-g particle moving in the xy plane varies in time a
ID: 2058639 • Letter: T
Question
The vector position of a 3.50-g particle moving in the xy plane varies in time according to r rightarrow = (3i^ + 3j^ )t + 2j^ t2,where t is in seconds and r rightarrow is in centimeters. At the same time, the vector position of a 5.50-g particle varies asr2 rightarrow = 3i^ - 2i^t2 - 6j^ t. At t = 2.50 s, determine (a) the vector position of the center of mass, (b) the linear momentum of the system, (c) the velocity of the center of mass,(d) the acceleration of the center of mass, and (e) the net force exerted on the two-particle system.Explanation / Answer
m1 = 3.5 g
r1 = 3ti+(3t+2t^2)j
m2 = 5.5 g
r2 = (3-2t^2)i -6tj
a)rcom = (m1r1 + m2r2)/(m1+m2)
= {3.5[3ti+(3t+2t^2)j]+5.5[(3-2t^2)i -6tj]}/(3.5+5.5)
= [(10.5t+16.5-11t^2)i + (10.5t+7t^2-33t)j]/9
= [(10.5t+16.5-11t^2)i + (7t^2-22.5t)j]/9
at t=2.5 s
rcom = -2.89i -1.39j
b)
vcom = d(rcom)/dt
= d([(10.5t+16.5-11t^2)i + (7t^2-22.5t)j]/9)/dt
= [(10.5-22t)/9]i + [(14t-22.5)/9]j
linear momentum of system = linear momentum of com
pcom = mcom*vcom
= 9*[(10.5-22t)/9]i + [(14t-22.5)/9]j
= [10.5-22t]i + [14t-22.5]j
at t=2.5s
pcom = -44.5i + 12.5j
c)
vcom =[(10.5-22t)/9]i + [(14t-22.5)/9]j
at t=2.5 s
vcom = -4.94i + 1.39j
d)
acom = d(vcom)/dt
=d([(10.5-22t)/9]i + [(14t-22.5)/9]j)/dt
= -22/9i + 14/9j</p>
at t=2.5s
acom = -2.44i + 1.56j
e)
force on system = force on com
Fcom = mcom*acom
= 9*( -2.44i + 1.56j)
= -22i + 14j
at t =2.5s
F on system =-22i + 14j
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