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[From: ] [author: ] [Date: 12-04-09] [Hit: ]
F delta(T) = 12.F (0.339 s = 12.F = 36.......
Please help me on a Linear momentum problem.THANKS
suppose a cart of mass m = 684 g is initial moving at speed v0 = 9.66 m/s. It strikes a spring attached to a fixed wall, comes to a stop, and bounces back at speed vf = 8.28 m/s. Assume the track is frictionless.
a) Find the magnitude of the impulse exerted by the spring on the cart.
b) If the cart was in contact with the spring for 339 ms, find the average force exerted by the spring on the cart.

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impulse equals change in momentum
the speed goes from +9.66 to -8.28 for a change of 17.94 m/s
mass is 0.684 kg
so
F delta(T) = m delta(V)
F delta(T) = 0.684 (17.94) = 12.27 kg*m/s or 12.27 N*s
b)
F delta(T) = 12.27
F (0.339 s = 12.27 N*s
F = 36.2 N
1
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