Two objects of masses m1 and m2 fall from the height respectively. The ratio of the magnitude of their momenta when they hit the ground?
M1/M2√h1/h2
(M1/M2) √(h1/h2)
To solve this problem, we need to use the principle of conservation of mechanical energy. When the two objects fall from a height, they both have potential energy, which gets converted into kinetic energy as they accelerate due to gravity.
Let's first assume that there is no air resistance, which means both objects experience the same acceleration due to gravity (g = 9.8 m/s^2).
The potential energy (PE) of an object of mass m at a height h is given by the formula PE = mgh. Since the objects fall from the same height, the potential energy for both objects is proportional only to their masses.
PE₁ = m₁gh
PE₂ = m₂gh
Now, when an object falls freely, its potential energy gets converted into kinetic energy (KE) when it reaches the ground. The kinetic energy depends on the mass and velocity of the object and is given by the formula KE = 1/2 mv^2.
Using the principle of conservation of mechanical energy, we can equate the potential energy to the kinetic energy:
m₁gh = 1/2 m₁v₁² (Equation 1)
m₂gh = 1/2 m₂v₂² (Equation 2)
Here, v₁ and v₂ are the final velocities of the two objects just before they hit the ground.
Now, to find the ratio of their momenta, we need to calculate their momenta first. The momentum (p) of an object is given by the product of its mass (m) and velocity (v):
p₁ = m₁v₁
p₂ = m₂v₂
Dividing Equation 1 by Equation 2, we get:
(m₁gh)/(m₂gh) = (1/2 m₁v₁²)/(1/2 m₂v₂²)
(m₁/m₂) = (v₁²/v₂²)
Taking square roots on both sides, we have:
(sqrt(m₁/m₂)) = (v₁/v₂)
Therefore, the ratio of the magnitude of their momenta when they hit the ground is the square root of the ratio of their masses:
p₁/p₂ = sqrt(m₁/m₂)
This means that the ratio of their momenta is equal to the square root of the ratio of their masses.