They have the same velocity because their displacements (shortest line from point A to point B, which is a straight line) are the same and they meet at the same time.
Answer:
U = 1 / r²
Explanation:
In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related
F = - dU / dr
this derivative is a gradient, that is, a directional derivative, so we must have
dU = - F. dr
the esxresion for strength is
F = B / r³
let's replace
∫ dU = - ∫ B / r³ dr
in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product
let's evaluate the integrals
U - Uo = -B (- / 2r² + 1 / 2r₀²)
To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)
U = B / 2r²
we substitute the value of B = 2
U = 1 / r²
Given
m1(mass of the first object): 55 Kg
m2 (mass of the second object): 55 Kg
v1 (velocity of the first object): 4.5 m/s
v2 (velocity of the second object): ?
m3(mass of the object dropped): 2.5 Kg
The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:
Pa= Pb
Where Pa is the momentum before collision and Pb is the momentum after collision.
Now applying this law for the above problem we get
Momentum before collision= momentum after collision.
Momentum before collision = (m1+m2) x v1 =(55+5)x 4.5 = 270 Kgm/s
Momentum after collision = (m1+m2+m3) x v2 =(55+5+2.5) x v2
Now we know that Momentum before collision= momentum after collision.
Hence we get
270 = 62.5 v2
v2 = 4.32 m/s
You haven't told us anything about the detectors being used. We don't know how the sensitivity of the detector is related to the total number of photons absorbed, and we don't even know whether you and your friend are both using the same type of detector.
All we can do, in desperation, is ASSUME that the minimum time required to just detect a star is inversely proportional to the total number of its photons that strike the detector. That is, assume . . .
(double the number of photons) ===> (detect the source in half the time) .
-- The intensity of light delivered to the prime focus of a telescope is directly proportional to the AREA of its objective lens or mirror, which in turn is proportional to the square of its radius or diameter.
So your telescope gathers (0.18/0.05)² = 12.96 times as much light as your friends telescope does.
-- So we'd expect your instrument to detect the same star in
(119.5 min) / (12.96) = <em>9.22 minutes .</em>
We're simply comparing the performance of two different telescopes as they observe the same object, so the star's magnitude doesn't matter.
