Change in momentum of different masses

AI Thread Summary
Change in momentum is directly related to mass and velocity, defined by the equation M(v-u) or impulse (F x time). When comparing a mass of 4m to a mass of m pushed with the same force, the change in momentum remains constant for both due to the relationship between mass and velocity. Specifically, if the mass increases by four times, the resulting change in velocity decreases by four times, keeping the overall change in momentum equal. This illustrates that for a given impulse, the change in momentum does not depend on the mass itself but rather the force applied over time. Understanding this relationship clarifies how mass affects momentum when a constant force is applied.
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Homework Statement


How is change in momentum related to mass? Will the change in momentum for a mass of 4m being pushed with constant force be same as the change in momentum for a mass of m being pushed with same force?

Homework Equations


The Attempt at a Solution



I know Change of momentum (i.e Impulse) is M(v-u) which is also equal to F.time. But that would suggest change in momentum for mass of $m is more than that of m. Is that right?
 
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Momentum is mass X velocity, or change in momentul is mass X change in velocity.
For a given impulse, F X Time, the change in momentum will be the same. If the mass is increased by 4 times, then the change in velocity will decrease by 4 times.
 
barryj said:
Momentum is mass X velocity, or change in momentul is mass X change in velocity.
For a given impulse, F X Time, the change in momentum will be the same. If the mass is increased by 4 times, then the change in velocity will decrease by 4 times.

Yup, that was my thought. Wanted to get it cleared. Thanks!
 
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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