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

Mass of block m = 150g = 0.15kg

Ball slide from h =2.44m

Taking g =10m/s²

let velocity of ball at bottom is v.

angular acceleration w=v/r

and I(moment of inertia of ball) = 2mr²/3

Using energy conservation

Total energy at top

u (potential energy)= mgh

Total energy at bottom

K(kinetic energy) = mv²/2 +Iw²/2

= (mv² + 2mr²×(v/r) ²/3) /2 putting I and w

= 5mv²/^​6

Now total initial energy = final energy

mgh = 5mv²/6

√(6gh/5) =v

V = √(6×10×2.44/5)

V =5.41m/s

Now ball stick to block of mass m_b =280g = 0.280kg

Let after velocity of block and ball v_s​​​​​

Using momentum conservation

Initial = final

mv = (m +m_b) v_s as both stick

0.15×5.41 = (0.15+0.28) v_s

v_s = 1.88m/s

If they have to climb hill of height h = 0.23m their initial energy should be greater than final energy mgh

Final potential energy = m_total​​​​​​gh

= (0.15 +0.28) ×10×0.23

= 0.989J

Initial kinetic energy = mv²/2

=(0.15+0.28) (1.88) ²/2

= 0.758J which is less than 0.989J

So it will not climb hill.