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Chapter 9: Problem 34

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 two blocks must have the same mass for both to have the same speed after the collision.

Step by step solution

01

Apply conservation of momentum

Before the collision, initial momentum is \(m*v\) where \(v\) is the initial velocity and after the collision, final momentum is \(m*v1 + M*v1\), where \(v1\) is the final common speed. Applying the conservation of momentum gives: \(m \cdot v = (m+M) \cdot v1\)
02

Apply conservation of energy

Because the collision is elastic, the kinetic energy is also conserved. Before the collision, the initial kinetic energy is \(\frac{1}{2}m \cdot v^2\), and after the collision, the final kinetic energy is \(\frac{1}{2}m \cdot v1^2 + \frac{1}{2}M \cdot v1^2\). Applying the conservation of kinetic energy gives: \(\frac{1}{2}m \cdot v^2 = \frac{1}{2}m \cdot v1^2 + \frac{1}{2}M \cdot v1^2\)
03

Solve the equations

From the conservation of momentum, we get \(v1=v \cdot \frac{m}{m+M}\). Substituting the value of \(v1\) into the conservation of energy equation and solving that equation for \(m\), you get, \(m = M\), implying that the masses of the two blocks must be the same.

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