Why Do My Momentum Problem Solutions Differ from the Textbook Answers?

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The discussion centers on discrepancies between calculated answers for two physics problems involving momentum and force. In the first problem, the calculated force exerted on a roof by falling hailstones is approximately 2808 N, while the textbook states it should be 1900 N, suggesting that neglecting gravity yields a closer estimate. In the second problem, the participant struggles to derive the force exerted by a ball dropped from a height, noting that the textbook answer simplifies to mg, indicating an average force at terminal velocity. The participant recognizes that the lack of a specified relative velocity complicates their calculations. Overall, the thread highlights the importance of understanding underlying principles and assumptions in physics problems.
neelakash
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I did the following two questions of the same category.But the answer is not matching.Please see if I went wrong.

Homework Statement



1.During rainfall,2000 hailstones per square metre (radius 0.5 cm) falls onto a 10mx10x roof with velocity 20m/s.FInd the average force exerted onto the roof without rebound.Density of hailstones 900kg/m3

2.A ball of mass m is dropped onto the ground from a certain height.Colliding elastcally,it again rises to that height.Again it falls.What is the force exerted onto the grond in a long interval of time?

Homework Equations



I think in both cases we may use this standard equation:
P=F_ext+u(dM/dt) where u is the relative velocity

The Attempt at a Solution



(1) In problem (1),the total mass is 30*pi.You may check it.
So,total force exerted on the roof
F=F_floor+F_gravity
Taking magnitudes---
=u(dM/dt)+Mg
=20*30*pi+30*pi*g (where g is the acceleration due to gravty).
~2808N
whereas the book says it is 1900 N
I noted that if we neglect the gravity term,the answer (1885N) is close to 1900N

(2)Here what I got is

F=F_floor+F_gravity
Taking magnitudes,-
=u(dm/dt)+mg
=d(mu/dt)+mg
since the u has not been giveen,I could not proceed more.
However,they have provided the answer mg.
 
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OK,first part is clear.They referred to terminal velocity.Hence,the term "average".So,F_ext=0
 
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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