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

[neelakash]

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.

 

[neelakash]

OK,first part is clear.They referred to terminal velocity.Hence,the term "average".So,F_ext=0

 

[m2003]

In problem (1), it appears that you have correctly used the equation P = F_ext + u(dM/dt) to calculate the total force exerted on the roof. However, the discrepancy between your answer and the book's answer could be due to a few factors. First, the density of hailstones may not be exactly 900 kg/m3, which would affect the total mass and therefore the total force. Additionally, the velocity of the hailstones may not be exactly 20 m/s, which would also affect the total force. It may be helpful to double check your calculations and make sure all units are consistent.

In problem (2), your approach is correct in using the equation P = F_ext + u(dm/dt) to calculate the force exerted on the ground. However, as you noted, the relative velocity u is not given, which makes it difficult to solve for the force. It may be helpful to use the equation for elastic collisions, which relates the velocities of the two objects before and after the collision. This would allow you to solve for the relative velocity u and then plug it into the equation for force. Additionally, in a long interval of time, the force exerted on the ground would be the weight of the ball (mg) as it falls under the influence of gravity. This is likely why the book's answer is mg.