Momentum problem

[bluesky]

I did the following two questions of the same category.But the answer is not matching.Please see if I went wrong.

 

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?

 

I think in both cases we may use this standard equation:

P=F_ext+u(dM/dt) where u is the relative velocity

 

(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.

[bluesky]

A car with a gun and a man is rest on a frictionless floor.The total mass is 50m where m is the mass of a single shell.Now,the man fires each shell with a muzzle speed 200 m/s.and the car recoils.What is the speed aftyer the 2nd time firing?

 

R_CM=[1/(50m)][49m*r_cf+m*r_sf]

(d/dt)R_CM=[1/(50m)] [49m*v_cf+m* v_sf ]=0

 

[v_sc=v_sf-v_cf where v_sc,v_sf,v_cf are velocity of shell w.r.t. car,velocity of shell w.r.t. floor, velocity of car w.r.t. floor]

 

R_CM=[1/(50m)][49m*r_cf+m*r_sf]

 

(d/dt)R_CM=[1/(50m)] [49m*v_cf+m* v_sf ]=0

 

49m*v_cf+m*[ v_sc + v_cf ] =0

This gives,v_cf=-[v_sc/50 m/s

=-200/50 (i) m/s where (i) is the unit vector

 

v_sf=(49/50)v_sc

 

w.r.t the same frame,

{48m*v'_cf+m*[v'_sc+v'_cf]}+m*(49/50)v_sc=0

49v'_cf=-v'_sc-(49/50)v_sc

v'_cf=-200(1/50+1/49)(i)

 

 

Please check if I went wrong anywhere

[bluesky]

I think I am correct.However,I have noticed that even if you do not take the whole of the system in the second case,even then the problem can be done.Just we have to use a more familiar version of conservation of linear momentum and apply it over the region of interest.

 

I want to know to what extent it is justified to use two different systems in a SINGLE problem to have a unique result?MY intuition suggests when the results match there should be some deeper way of understanding the physics.

If the result does not match,...then...it is an accident?