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

To solve this problem, we will apply the principle of conservation of mechanical energy. The initial total mechanical energy of each block is the same and is entirely in the form of kinetic energy.

Let's denote:
- H: the height reached by the block on the longer track
- H1: the height at which the block on the shorter track leaves the track
- H2: the maximum height above the end of the track reached by the block on the shorter track
- θ: the angle of inclination of the incline (θ = 50.0°)
- v0: the initial speed of each block (v0 = 7.38 m/s)
- m: the mass of each block (consider them to be identical)

Now let's calculate the heights:

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

We can use the conservation of mechanical energy to equate the initial kinetic energy to the final gravitational potential energy of the block at height H.

Initial kinetic energy = Final gravitational potential energy

(1/2)mv0^2 = mgh

where g is the acceleration due to gravity (approximately 9.8 m/s^2). Rearranging the equation, we get:

h = (v0^2)/(2g)

Since the block slides upward until it reaches a maximum height H above the ground, the height h can be substituted with H:

H = (v0^2)/(2g)

Substituting the given values of v0 and g, we can find H.

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

Again, using the conservation of mechanical energy, we equate the initial kinetic energy to the sum of the final gravitational potential energy at height H1 and the maximum height H2.

Initial kinetic energy = Final gravitational potential energy at H1 + Final gravitational potential energy at H2

(1/2)mv0^2 = mgH1 + mgH2

Simplifying the equation, we get:

H1 + H2 = (v0^2)/(2g)

Substituting the given values of v0 and g, we can find H1 + H2.

By substituting the given values, you can now solve for (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.