Toy Cannon: Calculating the Projectile Speed

AI Thread Summary
A toy cannon projects a 5.30g rubber ball using a spring compressed by 5.00 cm, with a spring constant of 8.00 N/m and a friction force of 0.0320 N. The initial calculations led to a projectile speed of 1.17 m/s, which differed from the expected 1.40 m/s. The discrepancy arose from the incorrect application of the friction force over the total distance traveled in the barrel. Correcting the formula to account for friction only over the distance of spring compression yields the accurate speed. The discussion highlights the importance of precise calculations in physics problems.
Dorothy Weglend
Messages
247
Reaction score
2
A toy cannon uses a spring to project a 5.30g soft rubber ball. Spring is compressed 5.00 cm, k=8.00 N/m. Barrel is horizontal.

The ball moves 15.0 cm through the barrel, with friction force of 0.0320 N between thebarrel and the ball. With what speed does the projectile leave the barrel of the cannon?

I thought this problem was simple, but the answer 'in the back' is different from what I am getting, so I thought I would ask for some help.

I use

mv^2/2 = kx^2/2 - fk(x+d)

x = compression of spring, d = distance along the barrel, so

m = 5.32 * 10^-3 m,
k = 8 N/m
x = 0.05 m
k = 8 N/m
fk = 0.032 N
d = 0.15 m

With these figures, I get v = 1.17 m/s.

The answer in the back is 1.40 m/s.

Can anyone see where I am going wrong?

Dorothy
 
Physics news on Phys.org
Are you sure you are supposed to use ... - fk(x+d)?
if you use mv^2/2 = kx^2/2 - fk(x) instead, it will give a v of 1.40 m/s
I think the problem was trying to help you-it gave you the TOTAL distance the projectile has to travel in the barrel. I tried to make a diagram below, I'm not sure it worked.

Oh, and you meant m=5.30 x 10-3 kg right?

/\/\/ is the compressed spring
X is the projectile...|-0.15m-|
-------------
/\/\-X---->
-------------
 
Last edited:
That makes sense. Thanks. Yes, I used 5.30E-3.

Thanks again,
Dorothy
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top