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Sedbober [7]
3 years ago
9

Two pieces of clay are moving directly toward each other. When they collide, they stick together and move as one piece. One piec

e 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?
Physics
1 answer:
Wittaler [7]3 years ago
3 0

Answer:

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

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

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v

Put the value intro the formula

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

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

v=-0.5\ m/s

We need to calculate the total initial kinetic energy

Using formula of kinetic energy

K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2

Put the value into the formula

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

K.E_{i}=0.31875\ J

We need to calculate the total final kinetic energy

Using formula of kinetic energy

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

Put the value into the formula

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

K.E_{f}=0.1125\ J

We need to calculate the energy lost during the collision

Using formula of energy lost

energy\ lost=\dfrac{0.31875-0.1125}{0.31875}

energy\ lost=\dfrac{11}{17}\ J

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

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