Two pieces of clay are moving directly toward each other. When they collide, they stick together and move as one piece. One piece having mass 300 g is moving to the right at a speed of 1 m/s. The other piece has mass 600 g and is moving to the left at a speed of 0.75 m/s. What fraction of the total initial kinetic energy is lost during the collision?

Answer :

Answer:

The fraction of the total initial kinetic energy is lost during the collision is [tex]\dfrac{11}{17}\ J[/tex]

Explanation:

Given that,

Mass of one piece = 300 g

Speed of one piece = 1 m/s

Mass of other piece = 600 g

Speed of other piece = 0.75 m/s

We need to calculate the final velocity

Using conservation of energy

[tex]m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v[/tex]

Put the value intro the formula

[tex]300\times10^{-3}\times1+600\times10^{-3}\times(0.75)=(300\times10^{-3}+600\times10^{-3})v[/tex]

[tex]v=\dfrac{00\times10^{-3}\times1+600\times10^{-3}\times(-0.75)}{(300\times10^{-3}+600\times10^{-3})}[/tex]

[tex]v=-0.5\ m/s[/tex]

We need to calculate the total initial kinetic energy

Using formula of kinetic energy

[tex]K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2[/tex]

Put the value into the formula

[tex]K.E_{i}=\dfrac{1}{2}\times300\times10^{-3}\times1^2+\dfrac{1}{2}\times600\times10^{-3}\times(0.75)^2[/tex]

[tex]K.E_{i}=0.31875\ J[/tex]

We need to calculate the total final kinetic energy

Using formula of kinetic energy

[tex]K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2[/tex]

Put the value into the formula

[tex]K.E_{f}=\dfrac{1}{2}\times(300\times10^{-3}+600\times10^{-3})\times(-0.5)^2[/tex]

[tex]K.E_{f}=0.1125\ J[/tex]

We need to calculate the energy lost during the collision

Using formula of energy lost

[tex]energy\ lost=\dfrac{0.31875-0.1125}{0.31875}[/tex]

[tex]energy\ lost=\dfrac{11}{17}\ J[/tex]

Hence, The fraction of the total initial kinetic energy is lost during the collision is [tex]\dfrac{11}{17}\ J[/tex]

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