'H' = height at any time
'T' = time after both actions
'G' = acceleration of gravity
'S' = speed at the beginning of time
Let's call 'up' the positive direction.
Let's assume that the tossed stone is tossed from the ground, not from the tower.
For the stone dropped from the 50m tower:
H = +50 - (1/2) G T²
For the stone tossed upward from the ground:
H = +20T - (1/2) G T²
When the stones' paths cross, their <em>H</em>eights are equal.
50 - (1/2) G T² = 20T - (1/2) G T²
Wow ! Look at that ! Add (1/2) G T² to each side of that equation,
and all we have left is:
50 = 20T Isn't that incredible ? ! ?
Divide each side by 20 :
<u>2.5 = T</u>
The stones meet in the air 2.5 seconds after the drop/toss.
I want to see something:
What is their height, and what is the tossed stone doing, when they meet ?
Their height is +50 - (1/2) G T² = 19.375 meters
The speed of the tossed stone is +20 - (1/2) G T = +7.75 m/s ... still moving up.
I wanted to see whether the tossed stone had reached the peak of the toss,
and was falling when the dropped stone overtook it. The answer is no ... the
dropped stone was still moving up at 7.75 m/s when it met the dropped one.
Answer:
The mass of the second weight is approximately 0.477 kg
Explanation:
The given parameters are;
The acceleration experienced by the two weights = 3.8 m/s²
The mass of the first weight = 1.08 kg
The formula for the acceleration, a, of weights attached to a friction pulley, is given as follows;
Where;
a = The common acceleration of the two weights
g = The acceleration due to gravity = 9.81 m/s²
M = The mass of the first weight = 1.08 kg
m = The mass of the second weight
Therefore, we have;
The mass of the second weight = m ≈ 0.477 kg
The mass of the second weight ≈ 0.477 kg.
Answer:
V=15.3 m/s
Explanation:
To solve this problem, we have to use the energy conservation theorem:
the elastic potencial energy is given by:
The work is defined as:
this work is negative because is opposite to the movement.
The gravitational potencial energy at 2.5 m aboves is given by:
the gravitational potential energy at the ground and the kinetic energy at the begining are 0.
According to the <u>saving of the momentum</u> law, the total momentum <em>before </em>collision equals to the total momentum <em>after </em>collision
After collection, the 2 carts lock together, one body with a mass and velocity
from the definition of the momentum
thus
we calculate the velocity