Question 3
1 answer:
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1) Time in the air: 0.78 s
The motion of the ball is a projectile motion, which consists of two independent motions:
- A horizontal motion with constant horizontal velocity
- A vertical motion with constant downward acceleration of
(acceleration of gravity)
The initial vertical velocity is
where the negative sign means the direction is downward.
The vertical position of the ball is given by
where
h = 4.50 m is the initial heigth of the ball when it starts falling down
The ball reaches the ground when y = 0, so we have:
This is a second-order equation; solving for t, we get
t = -1.18 s
t = 0.78 s
We discard the negative solution since it has no physical meaning, so we can say that the ball spent 0.78 s in the air.
2) Horizontal distance: 1.83 m
For this second part of the problem, we just have to consider the horizontal motion of the ball.
As we said previously, the motion of the ball along the horizontal direction is a uniform motion with constant velocity, which is given by
where u = 3.05 m/s is the initial speed and the angle of projection.
For a uniform motion, we can use the following relationship between distance covered and velocity:
and substituting t = 0.78 s, we find the total distance travelled along the horizontal direction by the ball before reaching the ground:
Answer:
a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m
b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m
c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m
d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m
Explanation:
Given that;
height of the ramp h1 = 0.40 m
foot of the ramp above the floor h2 = 1.50 m
assuming R = 15 mm = 0.015 m
density of steel = 7.8 g/cm³
density of aluminum = 2.7 g/cm³
a) distance that the solid steel sphere sliding down the ramp without friction;
we know that
distance = speed × time
d = vt --------let this be equ 1
according to the law of conservation of energy
mgh₁ = mv²
v² = 2gh₁
v = √(2gh₁)
from the second equation; s = ut + at²
that is; t = √(2h₂/g)
so we substitute for equations into equation 1
d = √(2gh₁) × √(2h₂/g)
d = √(2gh₁) × √(2h₂/g)
d = 2√( h₁h₂ )
we plug in our values
d = 2√( 0.40 × 1.5 )
d = 1.55 m
Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m
b)
distance that a solid steel sphere rolling down the ramp without slipping;
we know that;
mgh₁ = mv² +
ω²
mgh₁ = mv² +
(
mR²) ω²
v = √( gh₁ )
so we substitute √( gh₁ ) for v and t = √(2h₂/g) in equation 1;
d = vt
d = √( gh₁ ) × √(2h₂/g)
d = 1.69√( h₁h₂ )
we substitute our values
d = 1.69√( 0.4 × 1.5 )
d = 1.31 m
Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m
c)
distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;
we know that;
mgh₁ = mv² +
ω²
mgh₁ = mv² +
(
mR²) ω²
v = √( gh₁ )
so we substitute √( gh₁ ) for v and t = √(2h₂/g) in equation 1 again
d = vt
d = √( gh₁ ) × √(2h₂/g)
d = 1.549√( h₁h₂ )
d = 1.549√( 0.4 × 1.5 )
d = 1.2 m
Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m
d) distance that a solid aluminum sphere rolling down the ramp without slipping.
we know that;
mgh₁ = mv² +
ω²
mgh₁ = mv² +
(
mR²) ω²
v = √( gh₁ )
so we substitute √( gh₁ ) for v and t = √(2h₂/g) in equation 1;
d = vt
d = √( gh₁ ) × √(2h₂/g)
d = 1.69√( h₁h₂ )
we substitute our values
d = 1.69√( 0.4 × 1.5 )
d = 1.31 m
Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m