A 970kg car is at the top of a 38m -long, 2.5 ∘ incline. Its parking brake fails and it starts rolling down the hill. Halfway down, it strikes and sticks to a 1280kg parked car.
Ignoring friction, what's the speed of the joined cars at the bottom of the incline?
What the first car's speed would have been at the bottom had it not struck the second car.
2 answers
homework
38m*sin2.5°=1.66m
1.66/2=.829m <- 'halfway down'
v=Sqrt(2*9.81*.829)=4.03m/s
970*4.03=(970+1280)*V V=1.733m/s
.5(2250)(1.73)^2+2250*9.8*.829=21646.5 J
21646.5=.5(2250)(V)^2 V=4.39m/s
The speed of the joined cars is 4.39m/s
Sqrt(2*9.8*1.66)=5.7m/s
The first cars speed would have been 5.7m/s
I hope that wasn't too confusing.
1.66/2=.829m <- 'halfway down'
v=Sqrt(2*9.81*.829)=4.03m/s
970*4.03=(970+1280)*V V=1.733m/s
.5(2250)(1.73)^2+2250*9.8*.829=21646.5 J
21646.5=.5(2250)(V)^2 V=4.39m/s
The speed of the joined cars is 4.39m/s
Sqrt(2*9.8*1.66)=5.7m/s
The first cars speed would have been 5.7m/s
I hope that wasn't too confusing.
1 answer