A 5.0 kg cart initially at rest ejects a 0.80 kg package backward at 6.0 m/s relative to ground. Remaining cart then experiences 1.4 N resistance for 3.0 s. (a) Find immediate cart speed. (b) Find resistive impulse and later speed. (c) Find total kinetic energy after ejection and explain its source. (d) Explain Newton's third law and when momentum conservation is approximate.

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Immediate cart speed 1.14 m/s forward; impulse −4.2 N s; later speed 0.143 m/s; post-ejection kinetic energy about 17.1 J. Remaining mass is 5.0 − 0.80 = 4.20 kg. With negligible external impulse during ejection, 0 = 4.20v + 0.80(−6.0), giving v = 1.14 m/s forward. Resistance impulse is FΔt = −1.4 × 3.0 = −4.2 N s. Initial post-ejection cart momentum is 4.80 kg m/s, so later momentum is 0.60 kg m/s and speed 0.60/4.20 = 0.143 m/s. Kinetic energy is ½(4.20)(1.1429²) + ½(0.80)(6.0²) = 2.74 + 14.4 = 17.1 J. Internal chemical or elastic energy supplies it; kinetic energy was not conserved. Cart and package exert equal opposite forces for equal times, giving opposite impulses. Those third-law forces act on different bodies and are internal to the combined system. If resistance gives appreciable impulse during the ejection interval, system momentum changes externally. Conservation is then approximate and must be justified by comparing external impulse with ejection momenta.

Rocket-cart ejection and resistance

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