Som CAR B Q7. A 1000 kg car A, is coasting (no brakes, no engine) down a hill 15

Question

Som CAR B Q7. A 1000 kg car A, is coasting (no brakes, no engine) down a hill 150 m high. When it is at the middle of the hill, (a) What is its total energy? (b) What is its potential energy? (c) What is its velocity? (d) What will be its velocity at the bottom of the hill? At the bottom of the hill it has an inelastic collision with a car B, half its mass. Car B climbs another hill in front of it from only the effects of the collision, How fast will car B take off if it was at rest initially (before the collision)? How far up the hill will car B go? Does car A come to a stop? Or does it continue moving? If it continues moving, in which direction will it move? If car A continues moving how high up the hill will it go, and up which hill? (e) (f) (g) (h)

Explanation / Answer

Part A

From energy conservation

Total energy at the middle point of hill will be equal to total energy at highest point

Total energy at highest point will be

TE = KE + PE

TE = 0.5*m*Vi^2 + m*g*h

Vi = 0 m/sec & h = 150 m

TE = 0 + 1000*9.8*150

TE = 1470000 J

So TE at middle point will also be = 1470000 J

Part B

PE at middle of hill = mg*(h/2)

PEm = 1000*9.8*150/2 = 735000 J

Part C

Since TE at middle point of hill is

TEm = KEm + PEm = 1470000

KEm = 1470000 - 735000 = 735000 J

KEm = 0.5*m*Vm^2 = 735000 J

Vm = sqrt (2*735000/1000)

Vm = 38.34 m/sec

Part D

at the bottom of hill

TE = KEb + PEb = 1470000 J

PEb = 0, since h = 0

KEb = 0.5*m*Vb^2 = 1470000

Vb = sqrt (1470000*2/1000)

Vb = speed at the bottom of hill = 54.22 m/sec

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