Answer:
<u><em>First Reaction:</em></u>
=> 
<u><em>Second Reaction:</em></u>
=> 
<u><em>Combined Reaction:</em></u>
=> 
Answer:
h = 2.087 m
Explanation:
Given
m₁ = 3 kg
v₁ = 20 m/s
m₂ = 2 kg
v₂ = - 14 m/s
In a completely inelastic collision the colliding objects stick together after the collision and move together as a single object.
In the given problem, lets assume that the balls of putty are initially moving along the y axis, upward direction being the positive y direction. And the collision occurs at the origin of the coordinate system.
We can apply the equation
vs = (m₁*v₁ + m₂*v₂) / (m₁ + m₂)
⇒ vs = (3 kg*20 m/s + 2 kg*(- 14 m/s)) / (3 kg + 2 kg)
⇒ vs = 6.4 m/s (↑)
To calculate the maximum height h attained by the combined system of two balls of putty after the the collision, we use the expression for linear motion under gravity:
vf² = vi² - 2*g*h
where
vf = 0 m/s
g = 9.81 m/s²
vi = vs = 6.4 m/s
finally we get h:
h = vi² / (2*g)
⇒ h = (6.4 m/s)² / (2*9.81 m/s²) = 2.087 m
Answer:
Option C. The force between them would be 4 times larger than with the
initial masses.
Explanation:
To know which option is correct, we shall determine the force of attraction between the two masses when their masses are doubled. This can be obtained as follow:
From:
F = GMₐM₆/ r²
Keeping G/r² constant, we have
F₁ = MₐM₆
Let the initial mass of both objects to be m
F₁ = MₐM₆
F₁ = m × m
F₁ = m²
Next, let the masses of both objects doubles i.e 2m
F₂ = MₐM₆
F₂ = 2m × 2m
F₂ = 4m²
Compare the initial and final force
Initial force (F₁) = m²
Final (F₂) = 4m²
F₂ / F₁ = 4m² / m²
F₂ / F₁ = 4
F₂ = 4F₁ = 4m²
From the above illustrations, we can see that when the mass of both objects doubles, the force between them would be 4 times larger than with the
initial masses.
Thus, option C gives the correct answer to the question.
Yes,and because not everyone can wink and often that someone can only wink with one eye only
mass and inertia determine the amount of gravitational force an object experiences
