Question

A 1-kg ball is released from a height of 6 m, and a 2-kg ball is released from a height of 3 m. Air resistance is negligible as they fall. Which of the following statements about these balls are correct? (There could be more than one correct choice.)

A) As they reach the ground, the 1-kg ball will have more kinetic energy than the 2-ka ball because it was dropped from a greater height.
B) As they reach the ground, the 1-kg ball will be moving faster than the 2-kg ball.
C) Both balls will reach the ground with the same speed.
D) Both balls will reach the ground with the same kinetic energy.
E) Both balls will take the same time to reach the ground.

Answer 1

Answer:D

Step-by-step explanation:

Answer 2

To solve this problem we will apply the concepts related to the calculation of the given speed depending on the height and gravity, as well as the calculation of the kinetic energy and the time elapsed in the occurrence of the phenomena. These processes will be compared between the two objects to conclude the correct answers.

The velocity of the ball one with 1 kg is

`v_1 = √(2gh)\\v_1 = √(2(9.8)(6))\\v_1 = 10.84m/s`

The velocity of the ball two with 1 kg is

`v_2 = √(2gh)\\v_2= √(2(9.8)(3))\\v_2 = 7.66m/s`

The kinetic energy of ball one is

`KE = (1)/(2)mv^2 \\KE = (1)/(2) (1)(10.84)^2\\KE = 58.75J`

The kinetic energy of the ball two

`KE' = (1)/(2) mv^2\\KE' = (1)/(2) (2)(7.66)^2\\KE' = 58.67J`

Kinetic energy are almost the same.

Time taken for the fall in free falling for ball one

`t_1 = \sqrt{(2g)/(h)}\\t_1 = \sqrt{(2*9.8)/(6)}\\t_1 = 1.8s`

Time taken for the fall in free falling of ball two

`t_2 = \sqrt{(2g)/(h)}\\t_2 = \sqrt{(2*9.8)/(3)}\\t_2 = 2.55s`

Time taken for both are different

Therefore the correct option are:

D and B.