A 990kg car is at the top of a 38m -long. 2.5 degree incline. Its parking brake

Question

A 990kg car is at the top of a 38m -long. 2.5 degree incline. Its parking brake fails and it starts rolling down the hill. Halfway down, it strikes and sticks to a 1270kg parked car. Ignoring friction, what's the speed of the joined cars at the bottom of the incline? Express your answer to two significant figures and include the appropriate units. What the first car's speed would have been at the bottom had it not struck the second car. Express your answer to two significant figures and include the appropriate units. upsilon =

Explanation / Answer

a)Conserving energy to find velocity of car just before collision,

mgdSin=0.5mv2

v=sqrt(gdSin)=2.85m/s

Conserving momentum,

2.85*990=(990+1270)*v

v=1.25m/s

Conserving energy for system,

0.5Mu2+MgdSin=0.5Mv2

Solving for v,

v=sqrt(u2+2gdSin)=4.2m/s

b)Conserving energy from top to bottom,

mgdSin=0.5mv2

d=38m

v=sqrt(gdSin)=4.0m/s

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