Answer:
a). [tex]H=2.45m[/tex]
b). [tex]H_{max}=1.94m[/tex]
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:
[tex]\frac{1}{2}*m*v_o^2=m*g*H[/tex]
Cancel the mass as a factor in each element in the equation
[tex]H=\frac{v_o^2}{2*g}[/tex]
[tex]H=\frac{(6.94m/s)^2}{2*9.8m/s^2}[/tex]
[tex]H=2.45m[/tex]
b).
The other lost some kinetic energy so, use a projectile motion to determine the total height for the other bock:
[tex]E_k=E_p[/tex]
[tex]E_k=m*g*H_1[/tex]
[tex]E_k=\frac{1}{2}*m*v_o^2-\frac{1}{2}*m*v^2[/tex]
[tex]m*g*H_1=\frac{1}{2}*m*(v_o^2-v^2)[/tex]
Solve to v'
[tex]v^2=v_o^2-2*g*H_1[/tex]
[tex]v=\sqrt{v_o^2-2*g*H_1}=\sqrt{(6.94m/s)^2-2*9.8m/s^2*1.25m}[/tex]
[tex]v=4.8m/s[/tex]
[tex]H_{max}=H_1+\frac{v^2*sin(50)}{2*g}=1.25m+\frac{(4.8m/s)^2*sin(50)}{2*9.8m/s^2}[/tex]
[tex]H_{max}=1.94m[/tex]
Answer:
(a) H = 1.41 m
(b) H₁ + H₂ = 1.35 m
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
