Math Is Fun Forum

yonski

Member

Registered: 2005-12-14

Posts: 67

Motion with variable mass

Hi there, I'm having some trouble with this one :-(

"A particle whose initial mass is m is projected upwards at time t=0 with speed gT, where T is constant  At time t its speed is u and its mass has increased to me^(t/T). If the added mass is at rest when it is acquired, show that

d/dt [mue^(t/T)] = -mge^(t/T) .

Deduce that the mass of the particle at its highest point is 2m."

I've derived the differential equation using the impulse-momentum principle, but i'm stuck on the second bit. When I solve the equation I get

u = gT[2e^(-t/T) - 1] .

So what I was planning to do was to find du/dt since the highest point that the particle reaches will occur when du/dt = 0. However with the equation i've found du/dt never does equal zero!

Please help!

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Student: "What's a corollary?"
Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."

yonski

Member

Registered: 2005-12-14

Posts: 67

Re: Motion with variable mass

Anyone?

---

Student: "What's a corollary?"
Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Motion with variable mass

why would it occur when du/dt = 0? u is the speed, so the highest point is going to be when the speed is 0, not the acceleration 0

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yonski

Member

Registered: 2005-12-14

Posts: 67

Re: Motion with variable mass

Oh yeah! Haha i knew i'd be making some stupid mistake. I hate it when that happens! Thanks

---

Student: "What's a corollary?"
Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."