The vector position of a 3.60 g particle moving in the xy plane varies in time a
ID: 1373207 • Letter: T
Question
The vector position of a 3.60 g particle moving in the xy plane varies in time according to
The vector position of a 3.60 g particle moving in the xy plane varies in time according to F net = A Mu N a cm = cm/s2 (e) Determine the net force exerted on the two-particle system at t = 3.00. v cm = cm/s (d) Determine the acceleration of the center of mass at t = 3.00. p = g A cm/s (c) Determine the velocity of the center of mass at t = 3.00. r cm = cm (b) Determine the linear momentum of the system at t = 3.00. jt. (a) Determine the vector position of the center of mass at t = 3.00. it2 ? 6i ? 2r = 3r with arrow is in centimeters. At the same time, the vector position of a 5.95 g particle varies as jt2 where t is in seconds and right t + 2j i + 3 3r 1 =Explanation / Answer
a)
at t = 3s,
r1 = (3i + 3j)*3 + 2j *3^2
= 9i + 27 j
r2 = 3i - 2i *3^2 - 6j 3
= -15 i - 18 j
rcm = (m1*r1+m2*r2)/(m1+m2)
= (3.6*(9i + 27 j) + 5.95*(-15 i - 18 j))/(3.6+5.95)
= (-5.95 i -1.04 j) cm
|rcm| = 6.04 cm
b)
v1 = dr1/dt = (3i+3j) + 4*t j
at t = 3s,
v1 = 3i + 15 j
v2 = dr2/dt
= -4*t i - 6*j
at t = 3s,
v2 = -12i - 6j
Pcm = (m1*v1 + m2*v2)/(m1+m2)
= (3.6*(3i+15j) + 5.95*(-12i-6j))/(3.6+5.95)
= (-6.34i + 1.92j) g.cm/s
|Pcm| = 6.62 g.cm/s
c) Pcm = (m1+m2)*vcm
==> vcm = Pcm/(m1+m2)
= (-6.34i + 1.92j)/(3.6+5.95)
= (0.66i + 0.2j) cm/s
|vcm| = 0.69 cm/s
d) a1 = dv1/dt = 4j
a2 = dv2/dt = -4 i
acm = (m1*a1+m2*a2)/(m1+m2)
= (3.6*4j -4*5.95i/(3.6+5.95)
= (2.5i - 1.5j) cm/s^2
|acm| = 2.91 cm/s^2
e) Fnet = (m1+m2)*acm
= (3.6+5.95)*2.91
= 27.8 dyne
= 27.8 *10^-5 N
= 278 micro N
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