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Chapter 9: Problem 76

An object collides elastically with an equal-mass object initially at rest. If the collision isn't head-on, show that the final velocity vectors are perpendicular.

Short Answer

Expert verified
The final velocity vectors for the two objects are indeed perpendicular to each other when the collision is not head-on. This results from the principles of conservation of momentum and kinetic energy in elastic collisions.

Step by step solution

01

Set Up Initial Conditions

Before the collision, let the moving object have velocity \(v\) and the object at rest have velocity 0. Therefore the initial momentum (p) is \(m \cdot v + m \cdot 0 = mv\) and the initial kinetic energy (KE) is \(\frac{1}{2}m \cdot v^2\). Here \(m\) is the mass of the objects.
02

Set Up Post-Collision Conditions

Let the velocity of the first object after the collision be \(v_1\) and the velocity of the second object be \(v_2\). Due to conservation of momentum, \(m \cdot v_1 + m \cdot v_2 = mv\). Squaring this equation gives \(v^2 = v_1^2 + v_2^2 + 2 \cdot v_1 \cdot v_2\). And due to conservation of kinetic energy, \(\frac{1}{2}m \cdot v^2 = \frac{1}{2}m \cdot v_1^2 + \frac{1}{2}m \cdot v_2^2\), or \(v^2 = v_1^2 + v_2^2\).
03

Compare Equations

Comparing the two obtained equations from Step 2, we see that the only possibility is if \(2 \cdot v_1 \cdot v_2 = 0\). This implies that the velocities are perpendicular to each other, as the dot product between two perpendicular vectors is zero.

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Most popular questions from this chapter

You're an accident investigator at a scene where a drunk driver in a \(2400-\mathrm{kg}\) car has plowed into a \(1700-\mathrm{kg}\) parked car with its brake set. You measure skid marks showing that the combined wreckage moved \(25 \mathrm{~m}\) before stopping, and you determine a frictional coefficient of \(0.75\). What do you report for the drunk driver's speed just before the collision?

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