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 - kg ball because it was dropped from a greater height.
b) both balls will reach the ground with the same kinetic energy.
c) as they reach the ground, the 1 - kg ball will be moving faster than the 2 - kg ball.
d) both balls will take the same time to reach the ground.
e) both balls will reach the ground with the same speed.

Explanation:

Step1: Calculate kinetic energy using conservation of energy

By conservation of energy \(K = mgh\).
For the \(1 - kg\) ball: \(K_1=m_1gh_1=(1)(9.8)(6)=58.8\space J\).
For the \(2 - kg\) ball: \(K_2=m_2gh_2=(2)(9.8)(3)=58.8\space J\). So, \(K_1 = K_2\), which means option b is correct and option a is wrong.

Step2: Calculate speed using \(K=\frac{1}{2}mv^{2}\)

From \(K=\frac{1}{2}mv^{2}\), we can get \(v=\sqrt{\frac{2K}{m}}\).
For the \(1 - kg\) ball: \(v_1=\sqrt{\frac{2\times58.8}{1}}=\sqrt{117.6}\approx10.84\space m/s\).
For the \(2 - kg\) ball: \(v_2=\sqrt{\frac{2\times58.8}{2}}=\sqrt{58.8}\approx7.67\space m/s\). So, \(v_1>v_2\), option c is correct and option e is wrong.

Step3: Calculate time using \(h = v_0t+\frac{1}{2}gt^{2}\) (since \(v_0 = 0\))

From \(h=\frac{1}{2}gt^{2}\), we can get \(t=\sqrt{\frac{2h}{g}}\).
For the \(1 - kg\) ball: \(t_1=\sqrt{\frac{2\times6}{9.8}}\approx1.11\space s\).
For the \(2 - kg\) ball: \(t_2=\sqrt{\frac{2\times3}{9.8}}\approx0.78\space s\). So, \(t_1
eq t_2\), option d is wrong.

Answer:

B. Both balls will reach the ground with the same kinetic energy.
C. As they reach the ground, the 1 - kg ball will be moving faster than the 2 - kg ball.