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Chapter 9: Problem 30

A block of mass \(m\) undergoes a one-dimensional elastic collision with a block of mass \(M\) initially at rest. If both blocks have the same speed after colliding, how are their masses related?

Short Answer

Expert verified
The masses of the two blocks are equal, \(M = m\).

Step by step solution

01

Step 1. Establish Conservation of Momentum

According to the law of conservation of momentum, the total momentum before and after the collision should remain constant. Before the collision, only the first block is moving, with mass \(m\) and velocity \(v\). So the total momentum is: \(P_{initial} = mv + 0\). After the collision, both blocks are moving with the same speed, say \(v'\), so the total momentum is: \(P_{final} = mv' + Mv'\). Setting these two equal gives the equation \(mv = (m+M)v'\).
02

Step 2. Set up the condition of the Elastic Collision

It is given as an elastic collision. For elastic collisions, kinetic energy is also conserved. The initial kinetic energy is \(K_{initial} = \frac{1}{2}mv^2 + 0\), and the final kinetic energy is \(K_{final} = \frac{1}{2}mv'^2 + \frac{1}{2}Mv'^2\). Setting these two equal gives the equation \(\frac{1}{2}mv^2 = \frac{1}{2}(m+M)v'^2\).
03

Step 3. Solve the Equations for mass relations

Eliminate the velocity terms to solve for the mass relations. Dividing the second equation by the first equation cancels out the velocity terms and gives \(\frac{m}{m+M} = \frac{1}{2}\), which leads to the final answer \(M = m\).

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