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Chapter 9: Problem 82

A block of mass \(m_{1}\) undergoes a one-dimensional elastic collision with an initially stationary block of mass \(m_{2} .\) Find an expression for the fraction of the initial kinetic energy transferred to the second block, and plot your result for mass ratios \(m_{1} / m_{2}\) from 0 to 20.

Short Answer

Expert verified
The fraction of initial kinetic energy transferred to the second block after the collision is given by \(\frac{4m_{1}m_{2}}{(m_{1} + m_{2})^{2}}\).

Step by step solution

01

Conservation of Momentum

Denote the initial velocity of the first block as \(u\), and the final velocities of the first and second blocks as \(v_{1}\) and \(v_{2}\) respectively. The conservation of momentum principle states that the total momentum before collision equals to the total momentum after collision, which gives us: \(m_{1} u = m_{1} v_{1} + m_{2} v_{2}\).
02

Conservation of kinetic energy

The principle of conservation of kinetic energy states that the total kinetic energy before collision is equal to the total kinetic energy after the collision: \(\frac{1}{2} m_{1} u^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}\).
03

Solve for \(v_{2}\)

From step 1 and 2, we can solve the two equations simultaneously to eliminate \(v_{1}\), and we would find that : \(v_{2} = \frac{2m_{1}u + v_{2}(m_{1} - m_{2})}{m_{1} + m_{2}}\).
04

Transferred kinetic energy

The fractional kinetic energy transferred to the second block is given by: \(Fraction = \frac{m_{2}v_{2}^{2}}{m_{1}u^{2}}\). Substituting the expression for \(v_{2}\) from step 3 into this fraction expression, you can simplify it to: \(Fraction = \frac{4m_{1}m_{2}}{(m_{1} + m_{2})^{2}}\).
05

Plot result

With this expression, one can plot the fraction of the initial kinetic energy transferred to the second block as a function of the mass ratio \(m_{1} / m_{2}\), ranging from 0 to 20. This plot will show how the kinetic energy transfer changes with varying mass ratios.

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