Help with conservation of momentum question

1. Nov 6, 2004

jhson114

A circus cannon, which has a mass M = 2500 kg, is tilted at q = 35°. When it shoots a projectile at v0 = 30 m/s with respect to the cannon, the cannon recoils along a horizontal track at vcannon = 2 m/s with respect to the ground.

a) At what angle to the horizontal does the projectile move with respect to the ground?

b) What is the mass of the projectile?
HELP: Use conservation of momentum.
HELP: Since the system is initially at rest and since there are no external forces in the horizontal direction, the horizontal component of the momentum of the cannon must be equal and opposite to the horizontal component of the momentum of the projectile (in the ground reference frame!). So you need to find the momentum of the projectile in the ground reference frame.

I already solved a, and it is 37 degrees. but i just cant solve a.
any suggestions would be of great help.

2. Nov 6, 2004

jhson114

this is what i have:
momentum initial have to equal momentum final. therefore:
for X-coordinate: 0=(2500*2)+M*30cos37
for y-coordinate: 0=(2500*2)+M*30sin37
when i solve the x-coordinate equation, i get mass equal to 208.7,
which is incorrect.

Last edited: Nov 6, 2004
3. Nov 6, 2004

jhson114

any input will help.

4. Nov 6, 2004

Duarh

Momentum remains zero _with respect to the rest frame_. So your value for the x component of the projectile's velocity in the momentum equation in part b should be the one you calculated in part a. It's not equal to 30cos37 since the velocity with respect to the ground is not 30 (it was 30 with respect to the cannon), it's smaller. Just replace 30cos37 with 30cos(35)-2 and you should be fine.

5. Nov 6, 2004

jhson114

wow. thanks. it worked :) i have one more question. if the canon shoot horizontally and fires the same projectile at the same speed relative to the cannon. With what velocity does the cannon now recoil with respect to the ground? you would do the same thing as before, 0=(2500*x)+30cos(z)-x
for z, you would have to put an angle thats respect to the grond correct? would it be zero?

6. Nov 6, 2004

Duarh

basically, yes. but you shouldn't complicate life; you don't have to take any angle if you already know that the velocity of the projectile is 40 w. respect to the cannon and that it was shot horizontally, you know the x component of the velocity is the velocity itself=40 m/s with respect to the cannon. edit: there was a mistake here before; you know that 40 m/s is velocity with respect to the cannon, but you don't know the velocity with respect to the ground since that depends on the velocity of the cannon; that velocity is now (40 - Vc)

Last edited: Nov 6, 2004
7. Nov 6, 2004

jhson114

do you mean 30 m/s??

8. Nov 6, 2004

Duarh

oh, yup. somebody posted the same problem a short while ago with different values and I still have those in mind.

9. Nov 6, 2004

jhson114

so the equation will be: 0=(-2500*x) + (28*221.5)

221.5 being the mass of projectile

10. Nov 6, 2004

Duarh

:) well, it'd be very useful if you could confirm that yourself. Does it make sense? Do you understand how you get to it?

11. Nov 6, 2004

jhson114

it does make sense to me. total momentum has to be zero, and since canon is moving opposite direction as the projectile, it would be negative. momentum is mv, so mass is known, and velocity is unknown. relative to the ground projectile is moving at 28m/s with mass of 221.5. however, the answer i get is wrong. ahahah im confused

12. Nov 6, 2004

Duarh

Oh, the momentum expression isn't quite right. We can't substitute 28 into vx because the x direction velocity was 28 exactly because the cannon was recoiling with 2 m/s. vx should instead be (30-vc).

13. Nov 6, 2004

jhson114

oh isee how it is. now i understand. thank you very much!