Calculating the change of momentum in a system

AI Thread Summary
The change in momentum for the ball is calculated using the formula delta p = p(final) - p(initial), resulting in a value of -4 kg-m/s. This negative value indicates a reversal in direction, as the ball rebounds west after initially moving east. The assumption is made that the impact is head-on without any angles involved. It's emphasized that momentum is a vector quantity, requiring careful attention to direction when calculating changes. The problem, while seemingly simple, is part of a more complex worksheet, prompting verification of the solution.
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Homework Statement



A ball with momentum of 10 kg-m/s [E] hits an object and rebounds with momentum of 6 kg-m/s [W]. What was the change in momentum?

Homework Equations



delta p=p(final)-p(initial)

The Attempt at a Solution



delta p=p(final)-p(initial)
delta p=(6 kg-m/s)-(10 kg-m/s)
delta p= -4 kg-m/s

This problem seems so simple, and yet it's in the middle of a worksheet full of really hard problems, so I just wanted to make sure I'm doing it right.
 
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I am assuming that impact is head on i.e. there are no angles involved.

Do not forget that momentum is a VECTOR.
Hence if to the RHS one takes it as positive, it must be taken as negative when it reverses direction.
 
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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