What would be the resulting condition if Earth's axis was perpendicular to the Sun?
2 answers:
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1. +72.0 kg m/s
The momentum of an object is given by:
p = mv
where
m is the mass of the object
v is its velocity
Taking "to the right" as positive direction, for Elena we have
m = 60.0 kg is the mass
v = +1.20 m/s is the velocity
So, Elena's momentum is
2. -162.5 kg m/s
Here Madison is moving in the opposite direction of Elena (to the left), so her velocity is
v = -2.50 m/s
while her mass is
m = 65.0 kg
Therefore, her momentum is
3. -90.5 kg m/s
The total momentum of Elena and Madison is equal to the algebraic sum of their momenta; taking into account the correct signs, we have:
4. 0.72 m/s to the left
We can find the final speed of Elena and Madison by using the law of conservation of momentum. In fact, the final momentum must be equal to the initial momentum (before the collision).
The initial momentum is the one calculated at the previous step:
while the final momentum (after the collision) is given by
where
is Elena's mass
is Madison's mass
v is their final velocity
According to the law of conservation of momentum,
So we can find v:
and the direction is to the left, since the sign is negative.
Answer:
a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m
Explanation:
a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.
So y - y₀ = ut - 1/2gt²
y - y₀ = (v₀sinθ)t - 1/2gt²
substituting the values of the variables into the equation, we have
0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²
- 60 = 50t - 4.9t²
So, 4.9t² - 50t - 60 = 0
Using the quadratic formula to find t,
Since t cannot be negative, t = 11.29 s
b. We first need to find the impact vertical velocity component. Using
v = u - gt where u = initial vertical velocity component = v₀sinθ and t = 11.29 s and g = 9.8 m/s². So,
v = v₀sinθ - gt
= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s
= 50.01 m/s - 110.64 m/s
= -60.63 m/s
Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.
The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)
= √((-60.63 m/s)² + (72.77 m/s)²)
= √((3676 m²/s² + 5295.48 m²/s²)
= √(8971.48 m²/s²
= 94.72 m/s
The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°
So the impact velocity is 94.72 m/s at -39.8°
c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m