The vector position of a 3.80 g particle moving in the xy plane varies in time a
ID: 1435322 • Letter: T
Question
The vector position of a 3.80 g particle moving in the xy plane varies in time according to r_1 = (3i + 3j)t + 2jt^2 where t is in seconds and r is in centimeters. At the same time, the vector position of a 5.55 g particle varies as r_2 = 3i - 2it^2 - 6jt. Determine the vector position of the center of mass at t = 2.90. Determine the linear momentum of the system at t = 2.90. Determine the velocity of the center of mass at t = 2.90. Determine the acceleration of the center of mass at t = 2.90. Determine the net force exerted on the two-particle system at t = 2.90.Explanation / Answer
v1 = dr1/dt = 3i + 3j + 4j t = 3i + (3+4t) j
v2 = dr2/dt = -4ti - 6j
a1 = dv1/dt = 4 j
a2 = dv2/dt = -4i
a)
at t = 2.9
r1 = 3i
r cm = (m1*r1 + m2*r2)/(m1+m2)
rcm = ((3.8*(3it + 3jt + 2it^2)) + (5.55*(3i - 2it^2 - 6jt) ) / (3.8 + 5.55)
at t = 2.9 s
rcm = ( (3.8*3*2.9)i + (3.8*3*2.9)j + (3.8*2*2.9^2)j + (5.55*3)i - (5.55*2*2.9^2)i - (5.55*6*2.9)j) / (3.8 + 5.55)
rcm = -4.67i + 0.0434 j
(b)
at t= 2.9
v1 = 3i + (3+(4*2.9))j = 3i + 14.6 j
v2 = -(4*2.9)i - 6j = -11.6i - 6j
pcm = (m1*v1 + m2*v2)
Pcm = (3.8*10^-3*(3i + 14.6 j)) + (5.55*10^-3*(-11.6i - 6j))
Pcm = ((3.8*3)- (5.55*11.6))i + ((3.8*14.6)-(5.55*6)) j
Pcm = -53 i + 22.2 j
(C)
Vcm = Pcm/(m1+m2) = -5.67i + 2.4j
(d)
a1 = dv1/dt = 4 j
a2 = dv2/dt = -4i
acm = m1*a1 + m2*a2/(m1+m2)
acm = (3.8*4)j - (5.55*4)i / (3.8+5.55)
acm = -2.37 i + 1.62 j
(e)
anet = sqrt(2.37^2+1.62^2) = 2.87 cm/s^2 = 0.0287 m/s^2
Fnet = m*acm = (3.8+5.55)*10^-3*0.0287 = 287 u N
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