Object A of mass M is released from height H, whereas object B of mass 0.5M is released from height 2H. What is the ratio of the velocity of object A to that of object B immediately before they hit the ground?
A. 1 : 2
B. 1.41 : 1
C. 1 : 1.41
D. 1 : 4
E. 1 : 1
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(1/2)v^2 = g h
mass cancels
To determine the ratio of the velocities of object A and object B at the moment they hit the ground, we can use the law of conservation of energy.
According to the law of conservation of energy, the initial potential energy of the objects will be converted into the sum of their final kinetic and potential energies.
The potential energy of an object of mass M at height H is given by the formula:
Potential energy = mass (M) x acceleration due to gravity (g) x height (H)
So, the potential energy of object A at height H is:
Potential energy of A = M x g x H -----(1)
Similarly, the potential energy of object B at height 2H (twice the height of object A) is:
Potential energy of B = 0.5M x g x (2H) = M x g x H -----(2)
Since objects A and B are released from rest, their initial kinetic energies are zero.
When the objects hit the ground, all of their potential energy will be converted into kinetic energy. The kinetic energy of an object is given by the formula:
Kinetic energy = (1/2) x mass x velocity^2
Let's assume the final velocities of objects A and B are VA and VB, respectively. The kinetic energy of object A will be:
Kinetic energy of A = (1/2) x M x VA^2 -----(3)
And the kinetic energy of object B will be:
Kinetic energy of B = (1/2) x (0.5M) x VB^2 = (1/8) x M x VB^2 -----(4)
According to the law of conservation of energy, the sum of the final kinetic energy and potential energy of each object must be equal to the initial potential energy:
Potential energy of A + Kinetic energy of A = Potential energy of B + Kinetic energy of B
Substituting the expressions from above, we get:
M x g x H + (1/2) x M x VA^2 = M x g x H + (1/8) x M x VB^2
Canceling out the common terms, we get:
(1/2) x VA^2 = (1/8) x VB^2
Now, let's solve the equation for the ratio VA/VB:
VA^2 = (1/4) x VB^2
VA/VB = √(1/4) = 1/2
So, the ratio of the velocity of object A to that of object B immediately before they hit the ground is 1:2.
Therefore, the correct answer is A. 1 : 2.
To find the ratio of the velocities of objects A and B immediately before they hit the ground, we can use the principle of conservation of mechanical energy.
The potential energy of an object at height H is given by mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
The kinetic energy of an object is given by (1/2)mv^2, where m is the mass of the object and v is its velocity.
When an object is released from rest and falls freely, its potential energy at the initial height is converted entirely into kinetic energy just before it hits the ground.
Let's calculate the potential energy and kinetic energy for objects A and B.
For object A:
Potential energy at height H, U_A = M * g * H
Kinetic energy just before hitting the ground, K_A = (1/2) * M * v_A^2
For object B:
Potential energy at height 2H, U_B = 0.5M * g * 2H = M * g * H
Kinetic energy just before hitting the ground, K_B = (1/2) * 0.5M * v_B^2
Since potential energy is converted into kinetic energy, we have:
U_A = K_A
M * g * H = (1/2) * M * v_A^2
Similarly, for object B:
U_B = K_B
M * g * H = (1/2) * 0.5M * v_B^2
Simplifying these equations:
v_A^2 = 2 * g * H
v_B^2 = 4 * g * H
Taking the square root of both sides:
v_A = sqrt(2 * g * H)
v_B = sqrt(4 * g * H) = 2 * sqrt(g * H)
Now we can calculate the ratio of the velocities:
(v_A / v_B) = sqrt(2 * g * H) / (2 * sqrt(g * H))
= sqrt(2) / 2
Simplifying further:
(v_A / v_B) = 1 / sqrt(2)
The given answer choices are:
A. 1 : 2
B. 1.41 : 1
C. 1 : 1.41
D. 1 : 4
E. 1 : 1
From our calculation, we can see that the ratio of v_A to v_B is 1 : 1.41. Therefore, the correct answer is option C. 1 : 1.41.