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 = 6.94 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

(a) H = 1.41 m

(b) H₁ + H₂ = 1.35 m

Step-by-step explanation:

LONGER TRACK

To calculate the height H of the longer track, we use the equation of motion on an inclined plane:

V² = U² -2gH---------------------------------------------- (1)

H = (U²- V²)/ 2g------------------------------------------- (2)

Since the block came to rest at height H, it implies that the final velocity V =0

Vertical component of the Initial velocity U = 6.94Sin 50°

Substituting into (2)

H = (6.94Sin 50°)²/(20)

= 1.4131

= 1.41 m

SHORTER TRACK (First Motion)

For the shorter track, the velocity (Vf) of the block at the end of the track is calculated as thus:

Initial velocity , V₀= 6.94 m/s

The vertical component of the velocity is 6.94Sin 50°

From the Law of Equation:

V² = U² -2gH---------------------------------------------- (1)

Substituting into (1)

V² = (6.94 Sin 50⁰)² – (2 x10 x1.25)

= 28.2635 – 25

= 3.2635

Vf = √3.2635

= 1.8065m/s

= 1.81 m/s

SHORTER TRACK (2nd Motion)

The block flew off at the end of the track in a projectory motion as shown above. This implies that the velocity (Vf) will be tangential to the path of motion and inclined as 50⁰ to the horizontal.

The vertical component of Vf = 1.8065 Sin 50⁰

Initial Velocity U = 1.8065 Sin 50⁰

At the maximum height of trajectory, final velocity, V = 0

To calculate H₂, we deploy the equation of motion in equation (1)

Substituting our new values into (1), we have:

0 = (1.8065 Sin 50⁰)² – (2 x10) x H₂

H₂ = (1.8065 Sin 50⁰)²/ 20

= 0.09575 m

H₁ + H₂ = 1.25 + 0.09575

= 1.34575‬

= 1.35 m

Answer 2

a).
`H=2.45m`

b).
`H_(max)=1.94m`

Step-by-step explanation:

For the block that stays on the track, its maximal height is attained when all of the kinetic energy is converted to potential energy

a).

The height for the block on the longer track can by find using this equation:

`(1)/(2)*m*v_o^2=m*g*H`

Cancel the mass as a factor in each element in the equation

`H=(v_o^2)/(2*g)`

`H=((6.94m/s)^2)/(2*9.8m/s^2)`

`H=2.45m`

b).

The other lost some kinetic energy so, use a projectile motion to determine the total height for the other bock:

`E_k=E_p`

`E_k=m*g*H_1`

`E_k=(1)/(2)*m*v_o^2-(1)/(2)*m*v^2`

`m*g*H_1=(1)/(2)*m*(v_o^2-v^2)`

Solve to v'

`v^2=v_o^2-2*g*H_1`

`v=√(v_o^2-2*g*H_1)=√((6.94m/s)^2-2*9.8m/s^2*1.25m)`

`v=4.8m/s`

`H_(max)=H_1+(v^2*sin(50))/(2*g)=1.25m+((4.8m/s)^2*sin(50))/(2*9.8m/s^2)`

`H_(max)=1.94m`