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Chapter 9: Problem 47

Masses \(m\) and \(3 m\) approach at the same speed \(v\) and collide head-on. Show that mass \(3 m\) stops, while mass \(m\) rebounds at speed \(2 v\).

Short Answer

Expert verified
After the collision, the mass \(m\) rebounds with a speed of \(2v\), while the mass \(3m\) comes to a rest.

Step by step solution

01

Calculate initial total momentum

The total momentum before the collision is calculated as the sum of the momentums of each of the two entities involved in the collision. The momentum of an entity is given by the product of its mass and its velocity. Thus, the total initial momentum \(P_{i}\) is given by \(P_{i} = m*v + 3m*-v\), since the two entities are moving in opposite directions.
02

Calculate final total momentum

Let the final velocity of mass \(m\) be \(v_{1}\) and that of mass \(3m\) be \(v_{2}\). After the collision, the total final momentum \(P_{f}\) would be \(P_{f} = m*v_{1} + 3m*v_{2}\). But from the law of conservation of momentum, \(P_{i} = P_{f}\). This implies that \(m*v + 3m*-v = m*v_{1} + 3m*v_{2}\). Substituting \(0\) for \(v_{2}\) on the assumption that mass \(3m\) comes to rest after the collision, we get \(m*v_{1} = 2m*v\). Solving for \(v_{1}\), we find that \(v_{1} = 2v\), indicating that mass \(m\) rebounds at speed \(2v\).
03

Confirm the assumption

Now let's check if our assumption that mass \(3m\) comes to rest is correct. Substituting \(v_{1} = 2v\) into the equation \(m*v + 3m*-v = m*v_{1} + 3m*v_{2}\), we find that \(v_{2} = 0\), validating our assumption that mass \(3m\) indeed comes to rest after the collision.

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Most popular questions from this chapter

Two \(140-\mathrm{kg}\) satellites collide at an altitude where \(g=8.7 \mathrm{m} / \mathrm{s}^{2}\) and the collision imparts an impulse of \(1.8 \times 10^{5} \mathrm{N} \cdot\) s to each. If the collision lasts \(120 \mathrm{ms}\), compare the collisional impulse to that imparted by gravity. Your result should show why you can neglect the external force of gravity.

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