Question

A 50g ball is released from rest 1.0 above the bottom of thetrack

shown. It rolls down a straight 30o segment, thenback up
a parabolic segment whose shape is given byy=(1/4)x2 ,
where x and y are in m. How high will theball go on the right
before reversing direction and rolling backdown?

Answer 1

The ball, when rolling up the parabolic hill without slipping, converts its kinetic energy back into potential energy and reaches the same height of 1.0 m from which it was originally released, assuming no energy losses.

Step-by-step explanation:

To determine the vertical height reached by the ball when it rolls up the parabolic hill, we can use conservation of energy. Initially, the ball has potential energy due to its height and zero kinetic energy because it is released from rest. As the ball rolls down the incline and back up the opposite side, its potential energy is converted into kinetic energy and then back into potential energy at its highest point.

Using the equations 1/2mv² = mgh for the situation when the ball rolls without slipping, we can find its maximum height on the opposite side of the parabolic incline. For this, we also use the fact that the initial velocity (v) at the bottom can be calculated using the equation V = (2gh)², where h is the initial height of 1.0 m, g is the acceleration due to gravity (9.8 m/s²), and V is the velocity of the ball at the bottom of the incline.

First, we calculate the velocity at the bottom using the equation for a ball rolling without slipping: V = √(2gh) = √(2 * 9.8 m/s² * 1 m) = √(19.6 m²/s²) = 4.43 m/s. This velocity then becomes the initial velocity for the ball as it rolls back up the parabolic path.

The vertical height reached can then be found by rearranging the potential energy equation: h = v² / (2g). When we input our calculated velocity, we get h = (4.43 m/s)² / (2 * 9.8 m/s²) = 1.0 m, which is the same height it was released from, as it has simply converted all its kinetic energy back to potential energy without any losses.

Answer 2

The maximum height of the ball is 2 m.

Step-by-step explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

`y=(1)/(4)x^2`

We need to calculate the velocity

Using conservation of energy

`\Delta U_(i)+\Delta K_(i)=\Delta K_(f)+\Delta U_(f)`

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

`\Delta U_(i)=\Delta K_(f)`

Put the value into the formula

`mgh=(1)/(2)mv^2`

Put the value into the formula

`50*10^(-3)*9.8*1.0=(1)/(2)*50*10^(-3)* v^2`

`v^2=(2*50*10^(-3)*9.8*1.0)/(50*10^(-3))`

`v=√(19.6)`

`v=4.42\ m/s`

We need to calculate the maximum height of the ball

Using again conservation of energy

`(1)/(2)mv^2=mgh`

Here, h = y highest point

Put the value into the formula

`(1)/(2)*50*10^(-3)*(4.42)^2=50*10^(-3)*9.8* h`

`y=(0.5*(4.42)^2)/(9.8)`

`y=0.996\ m`

Put the value of y in the given equation

`y=(1)/(4)x^2`

`x^2=4*0.996`

`x=√(4*0.996)`

`x=1.99\ m\ \approx 2 m`

Hence, The maximum height of the ball is 2 m.