Question

A ball of mass 1 kg and another ball of mass 4 kg are dropped together from a 20 m tall building. After a fall of 10 m towards ground, their respective kinetic energies will be in ratio of (neglect air friction)
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Answer 1

Explanation:

Given A ball of mass 1 kg and another ball of mass 4 kg are dropped together from a 20 m tall building. After a fall of 10 m towards ground, their respective kinetic energies will be in ratio of

• So the height of the building is 20 m. Now two balls of mass 1 kg and mass 4 kg are dropped together from same height.
• We have v = u + at
• So u is the initial velocity, v is the final velocity, since u is zero for both, acceleration is same and also time is same.
• Therefore velocities will also be same.
• Now Kinetic energy K.E = ½ mv^2, since velocities are same, the ratio would be only dependent on mass.
•         So K.E1 / K.E2 = m1/m2
•                                  = ¼
•      So the ratio will be 1 : 4

Reference link will be

https://brainly.in/question/28874959

Answer 2

Given,

Mass of first ball = 1 Kg

Mass of a second ball = 4 Kg

Both balls are dropped together from a 20 m tall building.

To find,

The ratio of their respective kinetic energies after a fall of 10 m towards the ground.

Solution,

We can simply solve this numerical problem by using the following process:

Mathematically,

The kinetic energy of a body= 1/2 × (mass) × (velocity)^2

=> the kinetic energy of anybody depends only on its mass and velocity but not on its elevation or position or height

{Equation-1}

Now, according to the question;

Both balls are dropped from the same height, same initial velocity 0, and same acceleration (9.8 m/s^2), then after falling down the same distance, both balls will have the same velocity.{Statement-1}

Now, according to the equation-1 and statement-1;

The ratio of their respective kinetic energies after a fall of 10 m towards the ground

= {1/2 × (mass of the first ball) × (velocity of the first ball)^2}/{1/2 × (mass of the second ball) × (velocity of the second ball)^2}

= (mass of the first ball)/(mass of the second ball)

{according to the statement-1, both the balls have the same velocity}

= 1 Kg/4 Kg = 1:4

Hence, the ratio of their respective kinetic energies after a fall of 10 m towards the ground is equal to 1:4.