Physics

A 2.0-kilogram laboratory cart is sliding across a horizontal frictionless surface at a constant velocity of 4.0 meters per second east. What will be the cart’s velocity after a 6.0-newton westward force
acts on it for 2.0 seconds?

(1) 2.0 m/s east (3) 10. m/s east
(2) 2.0 m/s west (4) 10. m/s west

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  1. call east positive x
    initial velocity Vo = +4 m/s
    initial momentum = m Vo = +8 kg m/s
    Force = -6 newtons
    time of impulse = 2 sec.
    Force times time = change of momentum = (-6)(2) = -12 newton seconds
    so change of momentum is -12
    final momentum = 8 - 12 = -4 kg m/s
    final speed = -4/2 = -2 m/s or 2 m/s west

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  2. 2 west

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