Question

In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is launched vertically into the air at point T. In case B, a 1 kg block slides without friction down an identically shaped ramp and is also launched vertically at point T. Select the statement that best describes which object will go higher after launch, and why

Answer 1

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Step-by-step explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

Em₀ = U = m g y₁

final point. At the exit of the ramp

Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

Em₀ = Em_f

m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

v = w r

the moment of inertia of a sphere is

I =
`(2)/(5)` m r²

we substitute

m g (y₁ - y₂) = ½ m v² + ½ (
`(2)/(5)` m r²) (
`(v)/(r)`)²

m g h = ½ m v² (1 +
`(2)/(5)`)

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

mg h =
`(7)/(5)` (
`(1)/(2)` m v²)

v = √5/7 √2gh

This is the exit velocity of the vertical movement of the sphere

v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

starting point

Em₀ = U = mg y₁

final point

Em_f = K + U = ½ m v² + m g y₂

Em₀ = Em_f

mg y₁ = ½ m v² + m g y₂

m g (y₁ -y₂) = ½ m v²

v = √2gh

this is the speed of the box

v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

v² = v₀² - 2 g y

at the highest point v = 0

y = vo₀²/ 2g

for the sphere

y_sphere = 5/7 2gh / 2g

y_esfera = 5/7 h

for the block

y_block = 2gh / 2g

y_block = h

therefore the block reaches higher than the sphere

`\frac{y_(sphere)} {y_bolck} = 5/7`