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Consider a rocket (initial mass m o) accelerating from rest in free space. At first, as it speeds up, its momentum p increases, but as its mass m decreases p eventually begins to decrease. For what value of m is p maximum

Respuesta :

Answer:

m = m_o/e

Explanation:

The rocket equation is given as;

m(dv/dt) = -v_ex(dm/dt)

v_ex is the exhaust velocity

Now, formula for momentum is;

p = mv

Differentiating with respect to time, we have;

dp/dt = m'v + mv'

Where;

m' is mass rate

v' is rate of change in velocity

Since we are dealing with exhaust velocity and momentum(p) is Maximum when v = v_ex then mv' can also be written as: -m'v_ex.

Thus;

dp/dt = m'v - m'v_ex

dp/dt = m'(v - v_ex)

Now, from the rocket equation we dt will cancel out to give;

-dv/v_ex = dm/m

Integrating both sides;

-∫dv/v_ex = ∫dm/m

This gives;

-v/v_ex = In(m/m_o)

Since momentum(p) is Maximum when v = v_ex

Thus;

-v/v = In(m/m_o)

-1 = In(m/m_o)

e^(-1) = m/m_o

m = m_o(e^(-1))

m = m_o/e

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