Cons of Momentum: Calculating Speeds

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The discussion focuses on a physics problem involving a rocket's explosion and the calculation of the final speeds of its separated sections. The initial momentum equation is presented, leading to the expression for the final speed of the front section. Participants highlight the need for additional equations to solve for the unknowns, specifically the relationships between the masses and speeds of the rocket sections. The conversation emphasizes the importance of considering both the minimum and maximum final speeds of the front section based on varying mass conditions. The thread illustrates the complexities of momentum conservation in explosive scenarios.
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Homework Statement


A rocket with a mass M moves along an x-axis at the constant speed vi=40 m/s. A small explosion separates the rocket into a rear section (of mass m1) and a front section; both sections move along the x axis. The relative speed between the rear and front sections is 20 m/s. What are (a) the minimum possible value of final speed vf of the front section and (b) for what limiting value of m1 does it occur? (c) What is the maximum possible value of vf and (d) for what limiting value of m1 does it occur?







Homework Equations





The Attempt at a Solution




M(40m/s)=m1fv1f+m2fv2f

which simplifies to v2f=(40M+20mi)/M

Now there is three unknowns...i think its asking for a value though...
 
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Hi kristing! :smile:

You've left out some relevant equations:

m1 + m2 = M; v1 = v2 + 20.

Now try again! :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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