Question

A 150g ball slides down a perfectly smooth incline starting at a height of 2.44m, at rest. At the bottom of the incline is strikes and sticks to a 280g block. The block/ball combo are then free to move along a frictionless rollercoaster track. The first hill is only 23 cm high. Can it make it over this first hill? Defend your response with detailed analysis. If the combo CAN make it over the hill, how fast is it moving at the top of the hill?

Answer 1

This problem can be solved using conservation of energy

Initial potential energy = mgh = 0.15*g*2.44 = 3.59 J

converting this to kinetic energy at bottom of incline

1/2*0.15*v^2 = 3.59

v = 6.92 m/s

conserving momentum

0.15*6.92 = (0.15 + 0.28) vf (inelastic collision both bodies become one)

vf = 2.414 m/s

this energy = 1/2*0.43*vf^2 = 1.253 J

to check if it can climb 23 cm high hill

it potential energy at top of hill if it had climbed = 0.43*g*0.23 = 0.97 J

as it has more kinetic energy than this potential enrgy it can climb over the hill

for velocity at top of hill conserve energy

1.253 = 0.97 + 1/2*0.43*v^2

1/2*0.43*v^2 = 0.283

v = 1.147 m/s

Please upvote if you like the answer!!