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h12_6a

# h12_6a - Problem 1 a F=ma and the force supplied by both...

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Problem 1) a) F=ma and the force supplied by both engines is the same. Therefore car b will have a greater acceleration and hence greater velocity 1 m into the race. b) We know work is change in kinetic energy. Also, W=force*distance. The same amount of work is done on both cars and they traveled the same amount of distance to finish the race, so they have the same kinetic energy at the end of the race. c) From K=.5 mv 2 we can solve for momentum= mv and car A has larger momentum. d) We know the acceleration from F=ma. Car b has more acceleration and has traveled farther at 10 s. e) After 10 s, the two cars have not traveled the same distance, so the work done on each is different. From d=.5 at 2 we can find distance traveled. We then use W=Fd=K f -K i = K f . We find car B has more kinetic energy after 10 s. f) From K=.5 mv 2 we can solve for momentum= mv and both cars have equal momentum. This makes sense because force is change in momentum over a period of time. Problem 2) a) Before the explosion the velocity is zero. So the momentum is zero. b) From Newton’s third law, action=reaction so the forces are equal. c) From conservation of momentum, 0=m 1 v 1 + m 2 v 2 so p A,f =-500 kg*m/s

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d) Initially, p y,i =m blue v blue and p x,i =m red v red Afterwards, p y,f =(m blue + m red ) V y and p x,f =(m blue + m red ) V x . From conservation of momentum, we can solve for V y and V x . Then tan θ = V y / V x = p y,f / p x,f and just rank the thetas. Problem 3) From conservation of momentum, the velocity of the earth is 2.03*10 -23 m/s so we can safely say it’s negligible. Problem 4) (8.20) For this problem we need both the conservation of energy and conservation of momentum. We can break the problem up into 2 parts. In the first part, we calculate the
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h12_6a - Problem 1 a F=ma and the force supplied by both...

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