vdm/dt

  • For a body with “changing” mass but “constant” v, the second law is reduced to Fnet = vdm/dt
  • \begin{aligned}{{F}_{net}}&=\frac{d(mv)}{dt}\\&=v\frac{dm}{dt}\end{aligned}

vdmdtA

  • Consider a rocket that ejects propellant at 2000 m s-1, at the rate of 300 kg s-1.
  • We can imagine the ejected propellant to be a body moving leftward at constant speed of 2000 m s-1 with a mass that increases at the rate 300 kg s-1. This means that its leftward momentum is increasing at the rate of vdm/dt = 2000 × 300 = 600 000 kg m s-2.
  • This requires the rocket to exert a leftward force of 600 kN on the ejected propellant.
  • By Newton’s 3rd Law, we can infer that the thrust acting on the rocket is also 600 kN but rightward.

vdmdtB

  • Consider a water jet of speed 10 m s-1. Upon striking a vertical wall, the water loses all its horizontal speed and falls down vertically.
  • So we can imagine the water jet to be a body moving rightward at constant speed of 10 m s-1 but losing mass continuously as it hits wall. Supposing that 20 kg of water hits the wall per second, it implies the water jet is losing rightward momentum at the rate of vdm/dt = 10 × 20 = 200 kg m s-2.
  • This requires the wall to exert a leftward force of 200 N on the water jet.
  • By Newton’s 3rd Law, we can infer that the water jet pushes the wall with a rightward force of 200 N.

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