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A particle of mass m1 and speed v1 collides with a second
Chapter 4, Problem 4.48(choose chapter or problem)
A particle of mass \(m_{1}\) and speed \(v_{1}\) collides with a second particle of mass \(m_{2}\) at rest. If the collision is perfectly inelastic (the two particles lock together and move off as one) what fraction of the kinetic energy is lost in the collision? Comment on your answer for the cases that \(m_{1} \ll m_{2}\) and that \(m_{2} \ll m_{1}\).
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QUESTION:
A particle of mass \(m_{1}\) and speed \(v_{1}\) collides with a second particle of mass \(m_{2}\) at rest. If the collision is perfectly inelastic (the two particles lock together and move off as one) what fraction of the kinetic energy is lost in the collision? Comment on your answer for the cases that \(m_{1} \ll m_{2}\) and that \(m_{2} \ll m_{1}\).
ANSWER:Step 1 of 3
Consider a perfectly inelastic collision of two masses \(m_{1}\) and \(m_{2}\). The initial velocity of \(m_{1}\) is \(v_{1}\) and that of \(m_{2}\) is zero. After the collision, two particles lock together and move off as one.
Write the equation of initial momentum of two particles.
\(m_{1} v_{1}+m_{2}(0)=m_{1} v_{1}\)
The final momentum is \(\left(m_{1}+m_{2}\right) v\).
From the law of conservation of momentum,
\(\begin{aligned} m_{1} v_{1} & =\left(m_{1}+m_{2}\right) v \\ v & =\frac{m_{1}}{\left(m_{1}+m_{2}\right)} v_{1} \end{aligned}\)
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