Two paths lead to the top of a big hill. One is steep and direct, while the other is twice as long but less steep. How much more
1 answer:
Answer:
None.
Explanation:
- Gravitational potential energy, depends only of the mass on which the gravity is doing work, and the displacement produced by this force.
- As displacement depends only on the final and initial positions (in this case, the height of the hill), if we choose as our zero reference level the bottom of the hill, the change in gravitational potential energy will be as follows:
- As we can see, the only value of distance involved is the height of the hill , so it is independent of the distance travelled.
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Answer:
When the person is drunk the car travels 20.1 meters more than when the person is sober. The distance traveled by car is:
Person drunk: 28.1 m
Person sober: 8 m
Explanation:
The distance traveled can be found using the following equation:
Where:
v: is the speed = 90 km/h
t: is the time
When the person is sober, the time that takes to hit the brakes is 0.320 s, so we have:
And when the person is drunk, the time is 1.00 s, hence the distance is:
The distance traveled by the car when the person is drunk compared when the person is sober is:
Therefore, when the person is drunk the car travels 20.1 meters more than when the person is sober.
I hope it helps you!
Answer:
1) elastic shock, the velocity of the center of mass does not change
2) inelastic shock, he velocity of the mass center change
Explanation:
The position of the center of mass of your system is defined by
=
in this case we have two bodies
x_{cm} = (x₁m₁ + x₂ m₂)
the velocity of the center of mass is
x_{cm} = dx_{cm} / dt =
x_{cm} =
where M is the total mass of the system.
Therefore to answer this question we have to find the velocity of the body after the collision.
Let's use momentum conservation, where the system is formed by the two bodies, so that the forces have been internal during the collision.
Let's solve each case separately.
2) inelastic shock
initial instant. Before the crash
p₀ = m₁ v₀ + 0
final instant. After the collision with the cars together
p_f = (m₁ + m₂) v
p₀ = p_f
m₁ v₀ = (m₁ + m₂) v
v = v₀
let's find the velocity of the center of mass
M = m₁ + m₂
initial.
=
(m₁ vo)
final
=
( v) = v
v_{cm f} =
Let's find the ratio of the velocities of the center of mass
vcmf / vcmo =
therefore the velocity of the mass center change
1) elastic shock
initial instant.
p₀ = m₁ v₀
final moment
p_f = m₁ v_{1f} + m₂ v_{2f}
p₀ = p_f
m₁ v₀ = m₁ v_{1f} + m₂ v_{2f}
m₁ (v₀ - v_{2f}) = m₂ v_{2f}
in this case the kinetic energy is conserved
K₀ = K_f
½ m₁ v₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²
m₁ (v₀² - v_{1f}²) = m₂ v_{2f}²
m₁ (v₀ + v_{1f}) (v₀ - v_{1f}) = m₂ v_{2f}
we write our system of equations
m₁ (v₀ - v_{1f}) = m₂ v_{2f} (1)
m₁ (v₀ - v_{1f}) (v₀ + v_{1f}) = m₂ v_{2f}²
we solve the system
v₀ + v_{1f} = v_{2f}
we substitute and look for the final speeds
v_{1f} =
v_{2f} =
now let's find the velocity of the center of mass
initial
=
m₁ v₀
final
=
(m₁ v_{1f} + m₂ v_{2f} )
v_{cm f} = [
+
] v₀
v_{cm f} = ( m₁² - m₁m₂ +2 m₁m₂) v₂
v_{cm f} = (m₁² + m₁ m₂) v₀
let's look for the relationship
v_{cm f} / v_{cm o} = M
v_{cm f} / v_{cm o} = 1
therefore the velocity of the center of mass does not change
we see in either case the velocity of the center of mass does not change.