Question

After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 . To reach the rack, the ball rolls up a rampthat rises through a vertical distance of 0.53

Answer 1

Speed before hitting ramp: u = 2.90 m/s
Speed after reaching top of ramp: v = ? m/s
Vertical distance h= 0.53 m

The correct answer would incorporate that some momentum andkineticenergy*will* be lost when the ball hits ramp, assuming the ramp is a straight line. So this reduces starting speed to u' < u = 2.90m/s.
(I always thought they would make the ramps curved).

But here is the "simple" (wrong) answer where we just do a simple energy balance and assume:
KE_after + PE_after = KE_before

½mv² + mgh = ½mu²
v² + 2gh = u²
v = √(u²-2gh)

So in this case:
v = √(2.90²-2(9.81)(0.53))
v = √(8.41-10.40)
so that seems to suggest the ramp is too high to climb(?).

The minimum criticalinitial speedneeded so the ball could climb would bethe one that gives v=0, i.e.
u²=2gh, u = √(2gh) = √10.40 = 3.22m/s

So ramp can't be climbed unless u >=3.22 m/s