A ball of mass 522 g starts at rest and slides down a frictionless track, as shown below. It leaves the track horizontally, striking the ground a distance x = 0.81 m from the end of the track after falling a vertical distance h2 = 1.03 m from the end of the track. (a) At what height above the ground does the ball start to move?
(b) What is the speed of the ball when it leaves the track?
(c) What is the speed of the ball when it hits the ground?
Those were all wrong, I don't get it.
To solve this problem, we can use the principle of conservation of energy. The initial potential energy of the ball at height h1 is converted to kinetic energy as it slides down the track, and then to potential energy again as it falls vertically.
Let's calculate the answers step-by-step:
Step 1: Find the initial height h1 above the ground where the ball starts to move.
We can use the conservation of energy equation:
mgh1 = mv^2/2
where m is the mass of the ball, g is the acceleration due to gravity, and v is the velocity of the ball.
Since the ball is initially at rest, its initial velocity is 0. Therefore, the kinetic energy term on the right side of the equation is 0.
mgh1 = 0
h1 = 0
So, the ball starts to move at ground level, and h1 = 0.
Step 2: Find the speed of the ball when it leaves the track.
Using the conservation of energy again, we can equate the potential energy at height h2 to the kinetic energy just before the ball leaves the track.
mgh2 = mv^2/2
Solving for v:
v^2 = 2gh2
v = sqrt(2gh2)
Given that h2 = 1.03 m and g = 9.8 m/s^2:
v = sqrt(2 * 9.8 * 1.03)
v ≈ 4.39 m/s
So, the speed of the ball when it leaves the track is approximately 4.39 m/s.
Step 3: Find the speed of the ball when it hits the ground.
Since the ball falls a distance x = 0.81 m from the end of the track, we can calculate the time it takes to fall using the equation for free fall:
h2 = (1/2)gt^2
1.03 = (1/2) * 9.8 * t^2
t^2 = 1.03 / (1/2 * 9.8)
t^2 = 0.1051
t ≈ 0.324 s
Now, using this time and the horizontal distance x, we can find the horizontal component of the initial velocity using the equation:
x = vx * t
vx = x / t
vx = 0.81 / 0.324
vx ≈ 2.50 m/s
The vertical component of the velocity remains the same throughout the motion. Therefore, the speed of the ball when it hits the ground is given by the magnitude of the overall velocity:
v = sqrt(vx^2 + vy^2)
v = sqrt(2.50^2 + 4.39^2)
v ≈ 5.17 m/s
So, the speed of the ball when it hits the ground is approximately 5.17 m/s.
To solve this problem, we need to apply the principles of conservation of energy.
(a) To find the height above the ground where the ball starts to move, we need to calculate the potential energy at that point.
First, we need to find the height h1 where the ball starts to move. We can use conservation of energy to equate the potential energy at points h1 and h2, neglecting any losses due to air resistance.
Potential Energy at h1 = Potential Energy at h2
m * g * h1 = m * g * h2
Here, m is the mass of the ball (522 g) and g is the acceleration due to gravity (9.8 m/s^2).
Converting the mass to kilograms:
m = 522 g = 522/1000 kg = 0.522 kg
Substituting the values:
0.522 kg * 9.8 m/s^2 * h1 = 0.522 kg * 9.8 m/s^2 * 1.03 m
Simplifying the equation:
h1 = 1.03 m
Therefore, the ball starts to move at a height of 1.03 m above the ground.
(b) To find the speed of the ball when it leaves the track, we can use the principle of conservation of energy again, this time equating the potential energy at point h1 with the kinetic energy at that point.
Potential Energy at h1 = Kinetic Energy at h1
m * g * h1 = (1/2) * m * v^2
In this equation, v is the velocity of the ball when it leaves the track.
Substituting the values:
0.522 kg * 9.8 m/s^2 * 1.03 m = (1/2) * 0.522 kg * v^2
Simplifying the equation:
5.11 = 0.261 * v^2
Rearranging the equation:
v^2 = 5.11 / 0.261
Taking the square root of both sides:
v = √(5.11 / 0.261)
Calculating the value of v:
v ≈ 3.89 m/s
Therefore, the speed of the ball when it leaves the track is approximately 3.89 m/s.
(c) To find the speed of the ball when it hits the ground, we can use the principle of conservation of energy one more time, equating the potential energy at point h2 with the kinetic energy just before hitting the ground.
Potential Energy at h2 = Kinetic Energy at ground level
m * g * h2 = (1/2) * m * v^2
Using the same values as before:
0.522 kg * 9.8 m/s^2 * 1.03 m = (1/2) * 0.522 kg * v^2
Simplifying the equation:
5.11 = 0.261 * v^2
Rearranging the equation:
v^2 = 5.11 / 0.261
Taking the square root of both sides:
v = √(5.11 / 0.261)
Calculating the value of v:
v ≈ 3.89 m/s
Therefore, the speed of the ball when it hits the ground is also approximately 3.89 m/s.
m•g•h1=m•v²/2
v=sqrt(2•g•h1)=v(x)
t=x/v =x/v(x) =x/ sqrt(2•g•h1)
h2=g•t²/2=g• x²/ 2•2•g•h1=x/4•h1 =>
h1=x/4•h2=0.81/1.03 =0.79 m
(a) H=h1+h2= 1.03+0.79=1.82 m.
(b) v=v(x)=sqrt(2•g•h1)=
=sqrt(2•9.8•0.79) = 3.93 m/s.
(c)
t=x/v(x)=0.81/3.93 = 0.2 s.
v(y) = gt=9.8•0.2= 1.96 m/s.
v=sqrt{v(x)²+v(y)²} = sqrt(3.93²+1.96²) = 4.4 m/s.