479 Horizontal Motion: The horizontal component of velocity ... - Wuala

91962_05_R1_p0479-0512 6/5/09 3:53 PM Page 482
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R1–5. The boy jumps off the flat car at A with a velocity of
v¿ =4 ft>s relative to the car as shown. If he lands on the
second flat car B, determine the final speed of both cars
after the motion. Each car has a weight of 80 lb. The boy’s
weight is 60 lb. Both cars are originally at rest. Neglect the
mass of the car’s wheels.
v ¿ 4 ft/s
13
5 12
B
A
Relative Velocity: The horizontal component of the relative velocity of the boy with
respect to the car A is (y b>A) x = 4a 12 b = 3.692 ft>s. Thus, the horizontal
13
component of the velocity of the boy is
(y b) x = y A + (y b>A) x
A ; + B (y b) x = - y A + 3.692
[1]
Conservation of Linear Momentum: If we consider the boy and the car as a system,
then the impulsive force caused by traction of the shoes is internal to the system.
Therefore, they will cancel out. As the result, the linear momentum is conserved
along x axis. For car A
0 = m b (y b) x + m A y A
A ; + B 0 = a 60
32.2 b(y b) x + a 80
32.2 b(-y A)
[2]
Solving Eqs. [1] and [2] yields
y A = 1.58 ft>s
Ans.
(y b) x = 2.110 ft>s
For car B
m b (y b) x = (m b + m B) y B
A ; + B a 60
60 + 80
b(2.110) = a
32.2 32.2 b y B
y B = 0.904 ft>s
Ans.
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