Question

current attempt in progress
the two objects shown in the drawing are initially moving in opposite directions. they are on a horizontal frictionless surface, so the net external force acting on them is zero. the objects collide and move apart after the collision. the table shows five possible sets of initial and final momenta for the objects. which is the only set that could occur?

a b c d e

Explanation:

Step1: Calculate initial momentum for each option

For option a: \(P_{i}=+16\,\text{kg}\cdot\text{m/s}+(- 16\,\text{kg}\cdot\text{m/s}) = 0\,\text{kg}\cdot\text{m/s}\)
For option b: \(P_{i}=+16\,\text{kg}\cdot\text{m/s}+(-6\,\text{kg}\cdot\text{m/s})=+10\,\text{kg}\cdot\text{m/s}\)
For option c: \(P_{i}=+12\,\text{kg}\cdot\text{m/s}+(-2\,\text{kg}\cdot\text{m/s}) = +10\,\text{kg}\cdot\text{m/s}\)
For option d: \(P_{i}=+12\,\text{kg}\cdot\text{m/s}+(-28\,\text{kg}\cdot\text{m/s})=-16\,\text{kg}\cdot\text{m/s}\)
For option e: \(P_{i}=+6\,\text{kg}\cdot\text{m/s}+(-14\,\text{kg}\cdot\text{m/s})=-8\,\text{kg}\cdot\text{m/s}\)

Step2: Calculate final momentum for each option

For option a: \(P_{f}=+8\,\text{kg}\cdot\text{m/s}+12\,\text{kg}\cdot\text{m/s}=+20\,\text{kg}\cdot\text{m/s}\)
For option b: \(P_{f}=+4\,\text{kg}\cdot\text{m/s}+10\,\text{kg}\cdot\text{m/s}=+14\,\text{kg}\cdot\text{m/s}\)
For option c: \(P_{f}=-4\,\text{kg}\cdot\text{m/s}+14\,\text{kg}\cdot\text{m/s}=+10\,\text{kg}\cdot\text{m/s}\)
For option d: \(P_{f}=-8\,\text{kg}\cdot\text{m/s}+(-10\,\text{kg}\cdot\text{m/s})=-18\,\text{kg}\cdot\text{m/s}\)
For option e: \(P_{f}=-6\,\text{kg}\cdot\text{m/s}+14\,\text{kg}\cdot\text{m/s}=+8\,\text{kg}\cdot\text{m/s}\)

Step3: Apply law of conservation of momentum

Since the net - external force \(F_{ext}=0\), by the law of conservation of momentum \(P_{i} = P_{f}\). Only for option c, \(P_{i}=P_{f}=+10\,\text{kg}\cdot\text{m/s}\)

Answer:

c