Question

decide how the objects in the following table would be listed, if they were listed in order of increasing kinetic energy. that is, select \1\ next to the object with the lowest kinetic energy, select \2\ next to the object with the second lowest kinetic energy, and so on.
object
kinetic energy
a 4.9 g bullet moving at 680. m/s.
select
a 4.9 g bullet moving at 605. m/s.
select
a 6.4 g bullet moving at 680. m/s.
select
a 4.9 g bullet moving at 580. m/s.
select

Explanation:

Step1: Calculate kinetic energy formula

The formula for kinetic energy is \(K.E=\frac{1}{2}mv^{2}\), where \(m\) is mass in kg and \(v\) is velocity in m/s. First, convert mass from grams to kilograms (\(1\ g = 10^{-3}\ kg\)).

Step2: Calculate for each bullet

• For \(4.9\ g\) bullet (\(m = 4.9\times10^{-3}\ kg\)) moving at \(v = 680\ m/s\):

\(K.E_1=\frac{1}{2}\times4.9\times 10^{-3}\times(680)^{2}=\frac{1}{2}\times4.9\times10^{-3}\times462400 = 1133.88\ J\)

• For \(4.9\ g\) bullet (\(m = 4.9\times10^{-3}\ kg\)) moving at \(v = 605\ m/s\):

\(K.E_2=\frac{1}{2}\times4.9\times 10^{-3}\times(605)^{2}=\frac{1}{2}\times4.9\times10^{-3}\times366025 = 896.76\ J\)

• For \(6.4\ g\) bullet (\(m = 6.4\times10^{-3}\ kg\)) moving at \(v = 680\ m/s\):

\(K.E_3=\frac{1}{2}\times6.4\times 10^{-3}\times(680)^{2}=\frac{1}{2}\times6.4\times10^{-3}\times462400 = 1479.68\ J\)

• For \(4.9\ g\) bullet (\(m = 4.9\times10^{-3}\ kg\)) moving at \(v = 580\ m/s\):

\(K.E_4=\frac{1}{2}\times4.9\times 10^{-3}\times(580)^{2}=\frac{1}{2}\times4.9\times10^{-3}\times336400 = 823.18\ J\)

Step3: Order the kinetic energies

Comparing the values: \(K.E_4(823.18\ J)<K.E_2(896.76\ J)<K.E_1(1133.88\ J)<K.E_3(1479.68\ J)\)

Answer:

A \(4.9\ g\) bullet moving at \(580\ m/s\): \(1\)
A \(4.9\ g\) bullet moving at \(605\ m/s\): \(2\)
A \(4.9\ g\) bullet moving at \(680\ m/s\): \(3\)
A \(6.4\ g\) bullet moving at \(680\ m/s\): \(4\)