#427 The vector position of a 3.20 g particle

The vector position of a 3.20 g particle - Physics

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Free Chegg Question

The vector position of a 3.20 g particle moving in the xy plane varies in time according to

₁=

3+ 3

t+ 2t²

wheret is in seconds and  is in centimeters. At the same time, the vector position of a 5.00 g particle varies as

₂= 3− 2t²− 6t.

(a) Determine the vector position of the center of mass att=2.60.
_cm=cm

(b) Determine the linear momentum of the system att=2.60.
=g · cm/s

(c) Determine the velocity of the center of mass att=2.60.
_cm=cm/s

(d) Determine the acceleration of the center of mass att=2.60.
_cm=cm/s²

(e) Determine the net force exerted on the two-particle system att=2.60.
_net=µN

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Free Chegg Answer 1

 

Free Chegg Answer 2

m1 = 3.20 g,

r1 = 3t i + (3t +2t2) j
r1' = 3 i + (3 + 4t) j
r1'' = 4 j

t = 2.60s,

r1 = 7.8 i + 21.32 j,
r1' = 3 i +13.4 j,
r1'' = 4 j

m2 = 5.00 g,

r2 = (3 - 2t2) i - 6tj, r2' = -4t i - 6 j, r2'' = -4 i,

t = 2.50s, r2 = -10.52 i - 15.6 j, r2' = -10.4 i - 6 j, r2'' = -4 i,

a) rc = (m1r1 +m2r2)/(m1 + m2) =(-3.37 i - 1.19 j) cm
b) P = m1r1' + m2r2' =(-42.4 i + 12.88 j) g-cm/s
c) rc' = P/(m1 + m2) = (-10.6 i +1.57 j) cm/s
d) rc'' = (m1r1'' + m2r2'')/(m1 + m2) =(-2.44 i + 1.56 j) cm/s2
e) (m1 + m2) rc'' = (4 i - 4 j)g-cm/s2 = (-20 i + 12.8 j) * 10-5 N