Question

two balls of different masses are rolled down a hill at different velocities. they data of these events are recorded below.

ball 1 trial 1	ball 1 trial 2	ball 2 trial 1	ball 2 trial 2
velocity (m/s)	2	4	2	4

1. which ball in which trial had the most kinetic energy?

• required 1

ball 1 trial 1
ball 1 trial 2
ball 2 trial 1
ball 2 trial 2

Explanation:

Step1: Recall kinetic - energy formula

The formula for kinetic energy is $KE=\frac{1}{2}mv^{2}$, where $m$ is the mass in kilograms and $v$ is the velocity in m/s. First, convert the mass from grams to kilograms.

Step2: Calculate kinetic energy for Ball 1 Trial 1

$m_1 = 50g=0.05kg$, $v_1 = 2m/s$. Then $KE_1=\frac{1}{2}\times0.05\times2^{2}=\frac{1}{2}\times0.05\times4 = 0.1J$.

Step3: Calculate kinetic energy for Ball 1 Trial 2

$m_2 = 50g = 0.05kg$, $v_2 = 4m/s$. Then $KE_2=\frac{1}{2}\times0.05\times4^{2}=\frac{1}{2}\times0.05\times16 = 0.4J$.

Step4: Calculate kinetic energy for Ball 2 Trial 1

$m_3 = 100g=0.1kg$, $v_3 = 2m/s$. Then $KE_3=\frac{1}{2}\times0.1\times2^{2}=\frac{1}{2}\times0.1\times4 = 0.2J$.

Step5: Calculate kinetic energy for Ball 2 Trial 2

$m_4 = 100g = 0.1kg$, $v_4 = 4m/s$. Then $KE_4=\frac{1}{2}\times0.1\times4^{2}=\frac{1}{2}\times0.1\times16 = 0.8J$.

Answer:

Ball 2 Trial 2