two balls of masses in ratio 1:2 are dropped from the same height find :
a. the ratio between their velocities when they strike the ground and
b. the ratio of the forces acting on them during motion
(i) To determine the ratio between the velocities of the two balls when they strike the ground, we can use the principle of conservation of energy. When the balls are dropped from the same height, they both possess potential energy due to their positions, which is converted into kinetic energy as they fall. The conservation of energy can be expressed as:
Potential energy (PE) = Kinetic energy (KE)
Since the potential energy is directly proportional to the mass (m) and the height (h) while the kinetic energy is directly proportional to the mass (m) and the square of the velocity (v), we can write:
m1 * h = 1/2 * m1 * v1^2 (for ball 1)
m2 * h = 1/2 * m2 * v2^2 (for ball 2)
Given that the mass ratio between the balls is 1:2, we can express m2 in terms of m1 as m2 = 2 * m1:
m1 * h = 1/2 * m1 * v1^2
2 * m1 * h = 1/2 * 2^2 * m1 * v2^2
2 * h = v2^2
Taking the square root of both sides, we get:
sqrt(2 * h) = v2
Therefore, the ratio of velocities when the balls strike the ground is:
v2/v1 = sqrt(2 * h) / v1
(ii) The ratio of the forces acting on the balls during motion can be determined using Newton's second law of motion, which states that the force (F) acting on an object is equal to its mass (m) multiplied by its acceleration (a):
F = m * a
Since the balls are dropped from the same height, they experience the same acceleration due to gravity (g). Therefore, the ratio of forces acting on the balls can be expressed as:
F2/F1 = (m2 * a) / (m1 * a)
Since the acceleration (a) cancels out, we can simplify the expression:
F2/F1 = m2/m1 = (2 * m1) / m1 = 2
Therefore, the ratio of the forces acting on the balls during motion is 2:1.
v is the same
forces in ration 1:2
To find the ratio between the velocities of the two balls when they strike the ground, we can use the principle of conservation of energy. When the balls are dropped from the same height, the potential energy is converted into kinetic energy as they fall.
Let's denote the masses of the two balls as m1 and m2, with m1 being the smaller mass and m2 being twice as big as m1. Let's also denote the velocities of the balls when they strike the ground as v1 and v2, respectively.
a. The ratio between their velocities when they strike the ground:
The potential energy of an object can be calculated using the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.
Since both balls are dropped from the same height, the potential energy is the same for both balls: PE1 = PE2.
PE1 = m1gh (1)
PE2 = m2gh (2)
Since m2 is twice as big as m1, we can rewrite (2) as:
PE2 = (2m1)gh
According to the conservation of energy, the potential energy is converted into kinetic energy:
KE1 = 1/2 m1v1^2 (3)
KE2 = 1/2 m2v2^2 (4)
Since the potential energy is being converted into kinetic energy, we can equate the potential energy to the kinetic energy for each ball:
PE1 = KE1 (5)
PE2 = KE2 (6)
Substituting equations (1), (3), (5) into equation (6), we get:
m1gh = 1/2 m1v1^2
gh = 1/2 v1^2
v1^2 = 2gh
m2gh = 1/2 m2v2^2
2m1gh = 1/2 (2m1)v2^2
gh = 1/2 v2^2
v2^2 = 2gh
Taking the square root of both sides of the equations, we find:
v1 = √(2gh)
v2 = √(2gh)
Since the height (h) and acceleration due to gravity (g) are constant, we can conclude that:
v1 = v2
Therefore, the ratio between their velocities when they strike the ground is 1:1.
b. The ratio of the forces acting on them during motion:
The force acting on an object can be calculated using Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a): F = ma.
Since the balls are dropped from the same height and experience the same acceleration due to gravity, the ratio of the forces acting on them during motion is the same as the ratio of their masses:
F1/F2 = m1/m2
Substituting M1 = m2 and M2 = 2m1, we get:
F1/F2 = m1/(2m1) = 1/2
Therefore, the ratio of the forces acting on them during motion is 1:2.