Solving the Inelastic Collision of 3 Balls

AI Thread Summary
The discussion revolves around solving an inelastic collision problem involving three balls with given masses and initial velocities. The user sets up momentum equations in both x and y directions but initially struggles with the correct sign conventions for the velocities. After recalculating and correcting the signs, they find the correct initial velocity for Ball 2 to be 13.11 m/s. The user acknowledges a mistake in their setup regarding the direction of the velocities, which led to confusion. Ultimately, they successfully arrive at the correct answer after addressing the sign issues.
kk727
Messages
54
Reaction score
0

Homework Statement



2ur1lwy.png


Ball 1: 3kg, 7 m/s initially
Ball 2: 4kg, V m/s initially
Ball 3: 5kg, 10 m/s initially

Find V if the system collides inelastically and moves off at 2 m/s at 5 degrees above the horizontal.

Homework Equations



p + p = p + p (in both x and y directions)

The Attempt at a Solution



I set up the equations for momentum in both directions (sin and cosine). I ended up with 4v*sinθ = 13.7558 and 4v*cosθ = 50.608. I then solved for θ and got 15.2 degrees and an initial velocity for Ball 2 of 13.11 m/s. I don't believe this is the answer my teacher had in class.

My only question is, when setting up the problem, would you consider the velocity of Ball 3 to be negative? That's how I calculated it. I have no problems with the calculations, just the setting up of the problem. I didn't write all of my work down, but it was something like this:

(sin 15)(7)(3) + (sin θ)(v)(4) + (5)(-10)(sin 20) = (2)(sin5)(12)

And I simplified it, did the same for cosines, solved for θ, and then plugged into solve for V.

Am I way off?

:/
 
Physics news on Phys.org
(sin 15)(7)(3) should be negative, too - the vertical component is down.
 
Thank you! I realized I had my signs all wrong...stupid mistake! I figured it out and got the right answer!
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top