The drawing shows two frictionless inclines that begin at ground level (h = 0 m) and slope upward at the same angle θ. One track is longer than the other, however. Identical blocks are projected up each track with the same initial speed v0. On the longer track the block slides upward until it reaches a maximum height H above the ground. On the shorter track the block slides upward, flies off the end of the track at a height H1 above the ground, and then follows the familiar parabolic trajectory of projectile motion. At the highest point of this trajectory, the block is a height H2 above the end of the track. The initial total mechanical energy of each block is the same and is all kinetic energy. The initial speed of each block is v0 = 7.38 m/s, and each incline slopes upward at an angle of θ = 50.0°. The block on the shorter track leaves the track at a height of H1 = 1.25 m above the ground. Find (a) the height H for the block on the longer track and (b) the total height H1 + H2 for the block on the shorter track.

Question

The drawing shows two frictionless inclines that begin at ground level (h = 0 m) and slope upward at the same angle θ. One track is longer than the other, however. Identical blocks are projected up each track with the same initial speed v0. On the longer track the block slides upward until it reaches a maximum height H above the ground. On the shorter track the block slides upward, flies off the end of the track at a height H1 above the ground, and then follows the familiar parabolic trajectory of projectile motion. At the highest point of this trajectory, the block is a height H2 above the end of the track. The initial total mechanical energy of each block is the same and is all kinetic energy. The initial speed of each block is v0 = 7.38 m/s, and each incline slopes upward at an angle of θ = 50.0°. The block on the shorter track leaves the track at a height of H1 = 1.25 m above the ground. Find (a) the height H for the block on the longer track and (b) the total height H1 + H2 for the block on the shorter track.

Answer 1

Each Track

Vo = 7.38 m/s. @ 50o.
Xo = 7.38*cos50 = 4.74 m/s.
Yo = 7.38*sin50 = 5.65 m/s.

The Longer Track
a. hmax = (Y^2-Yo^2)/2g.
hmax = (0-31.9)/-19.6 = 1.63 m.

The Shorter Track
b. Y^2 = Yo^2 + 2g*d
Y^2 = 31.9 - 19.6*1.25 = 7.4
Y = 2.72 m/s = Velocity at 1.25 m.

hmax = 1.25 + (0-(2.72)^2/-19.6=1.63 m.

Answer 2

To find the height H for the block on the longer track and the total height H1 + H2 for the block on the shorter track, we can use the conservation of mechanical energy. The initial total mechanical energy, which is all kinetic energy, is the same for both blocks.

Let's begin by calculating the height H for the block on the longer track.

(a) Finding the height H for the block on the longer track:

Step 1: Determine the initial kinetic energy (KE) of each block:
The initial kinetic energy of each block is given by:
KE = (1/2) * m * v0^2

Given:
v0 = 7.38 m/s

Step 2: Calculate the initial kinetic energy (KE) for each block:
KE = (1/2) * m * v0^2

Step 3: Express the initial kinetic energy (KE) in terms of the height H:
KE = m * g * H

Step 4: Solve for H:
H = KE / (m * g)

(b) Finding the total height H1 + H2 for the block on the shorter track:

Step 1: Calculate the height H1:
H1 = 1.25 m (given)

Step 2: Determine the initial potential energy (PE) at the highest point of the parabolic trajectory:
The initial potential energy at the highest point is given by:
PE = m * g * H2

Step 3: Calculate the total mechanical energy (E) at the highest point:
The total mechanical energy (E) at the highest point is the sum of the initial kinetic energy (KE) and the initial potential energy (PE):
E = KE + PE = m * g * (H + H2)

Step 4: Express the total mechanical energy (E) in terms of the height H1:
E = m * g * (H1 + H2)

Step 5: Solve for H1 + H2:
H1 + H2 = E / (m * g)

Now, let's calculate the values:

Given:
v0 = 7.38 m/s
θ = 50.0°
H1 = 1.25 m

Step 2 (block on the longer track):
KE = (1/2) * m * v0^2
KE = (1/2) * m * (7.38 m/s)^2

Step 4 (block on the longer track):
H = KE / (m * g)

Step 1 (block on the shorter track):
H1 = 1.25 m (given)

Step 3 (block on the shorter track):
E = m * g * (H1 + H2)

Step 5 (block on the shorter track):
H1 + H2 = E / (m * g)

Now you can substitute the given values into the equations and solve for the desired heights.

Answer 3

To solve this problem, we need to consider the conservation of energy. The initial total mechanical energy of each block is the same and is all kinetic energy.

(a) To find the height H for the block on the longer track, we need to determine the maximum potential energy it reaches.

We can start by calculating the initial kinetic energy of the block on the longer track. Since the initial total mechanical energy is all kinetic energy, we can use the formula for kinetic energy:

Kinetic energy = 0.5 * mass * velocity^2

Given:
Initial speed (v0) = 7.38 m/s

Next, we need to determine the mass of the block. However, the mass is not given in the problem. So, we can apply the principle of inertia, which states that the mass of an object does not affect its motion on an inclined plane as long as there is no friction. Therefore, we can ignore the mass in this problem.

Now, let's calculate the initial kinetic energy:

Initial kinetic energy = 0.5 * velocity^2

Substituting the value of velocity (v0) into the equation:

Initial kinetic energy = 0.5 * (7.38 m/s)^2

Now, we have the initial kinetic energy. Since the total mechanical energy is conserved, this energy will be converted into potential energy at the highest point of motion on the longer track.

At the highest point, all of the initial kinetic energy will have been converted to potential energy. Therefore, the height H can be determined using the equation for potential energy:

Potential energy = mass * gravity * height

Since the mass cancels out in the calculation, we can write:

Potential energy = gravity * height

Simplifying the equation, we have:

height = Potential energy / gravity

Substituting the calculated initial kinetic energy into the equation:

height = (Initial kinetic energy) / gravity

Now, calculate the height H using the above equation.

(b) To find the total height H1 + H2 for the block on the shorter track, we need to determine the heights H1 and H2 separately.

We're given that the block on the shorter track leaves the track at a height H1 = 1.25 m above the ground. This means that at this point, all of the initial kinetic energy has been converted into potential energy.

Using the same equation for potential energy as before, we can determine the height H1.

Next, to calculate the height H2, we need to consider the parabolic trajectory of projectile motion. The highest point of this trajectory is reached when the vertical velocity component becomes zero. At this point, all of the potential energy has been converted into kinetic energy.

Using the equation for potential energy:

Potential energy = mass * gravity * height

Since the mass cancels out, we can write:

Potential energy = gravity * height

Set the potential energy at the highest point equal to the initial kinetic energy, and solve for height. This will give us H2.

Finally, calculate the total height H1 + H2 to find the answer.