Conservation Linear Momentum (? Book Error ?)

I think, the elevation of the barrel is 10 degree but the shot is not leaving the barrel at 10 deg. with the horizontal because of recoiling of the gun. The angle of projection is more than 10 degree with horizontal and hence Vbx = 200*sin(10^o) is not correct.

 

moreover the horizontal component is not sin@ but cos@.

Muzzle speed is relative to the berral

The book is giving correct answer.

 

And with this, you find the same result given by your book

v=20 ft/s
M= cannon mass
m= cannonball mass
V= cannon speed

Conservation of momentum in the x direction requires (absolute values):

mxv=MxV

which leads to:

V=[cos(10°)x200ft/sx10lb]/400lb=4.924 ft/s


Now… what about the vertical (i.e. y) v-component? The mxvxsin(10°) momentum seems to have no "counterpart"

 

Sorry ZZZZZ

During the process of firing, the cannon starts recoiling and acquirs a velocity [tex] v_c [/tex] and hence the velocity of ball relative to earth in horizontal direction after leaving the cannon will be [tex] v_b\cos \theta - v_c [/tex]

The vertical component of velocity will remains the same that is why the angle of projection will be greater then the angle of elevation of the barrel, as I gave in previous posting.

Now conserving the linear momentum

[tex] m(v_b\ cos\theta - v_c)\ = M v_c [/tex]

gives
[tex]v_c = \frac{m v_b cos\theta}{M + m}[/tex]
= 4.8 f/s negative is due to direction, arrange.

It is a well known problem asking about the direction of the ball leaving the gun.

 

Ahaaa, got it... I read the first post too quickly and I thought that 4.9240 ft/s was the answer in the book... actually zzzzzz means that I'm most of the time asleep and I should wake up before reading. Excuse me