Difficult Conservation question - HELP

  • Thread starter Thread starter Lorax
  • Start date Start date
  • Tags Tags
    Conservation
AI Thread Summary
The discussion revolves around a complex pool shot involving conservation of momentum and energy principles. A pool shark must sink a target ball at a 30-degree angle after colliding with it using a cue ball that initially travels at 1.00 m/s. The cue ball will bounce off the target ball at a 315-degree angle with a velocity of 0.75 m/s. To determine the target ball's velocity post-collision, the law of conservation of momentum is applied, requiring calculations based on the initial and final velocities of both balls. The conversation emphasizes breaking down the problem step by step to visualize and solve it effectively.
Lorax
Messages
41
Reaction score
0
63. A pool shark is forced to do a tricky shot in order to win a game. He needs to sink the target ball in the corner pocket at an angle of 30 degrees away from the collision location.

The Cue ball needs to bounce off of the other ball at an angle of 315 degrees with a velocity of of 0.75 m/s. If he gives a cue ball a velocity of 1.00 m/s @ ) degrees what is the velocity of the target ball after the collision?


--I'm just having a hard time visualizing this one, and if someone could help start me off that would be swell.
 
Physics news on Phys.org
Any help would be awesome.
 


This is definitely a difficult conservation question. Let's break it down step by step to try and understand it better. First, we have a pool shark who needs to sink the target ball in the corner pocket at a 30 degree angle away from the collision location. This means that the cue ball needs to hit the target ball at a specific angle and with a specific velocity in order to make the shot.

Next, we have the cue ball bouncing off the target ball at an angle of 315 degrees and with a velocity of 0.75 m/s. This information is crucial for us to solve the problem. We also know that the cue ball has an initial velocity of 1.00 m/s at an unknown angle.

To find the velocity of the target ball after the collision, we need to use the law of conservation of momentum. This law states that the total momentum before a collision is equal to the total momentum after the collision. In this case, the total momentum before the collision is the momentum of the cue ball (1.00 m/s) and the momentum of the target ball (unknown).

After the collision, the cue ball will still have a momentum of 1.00 m/s, but the target ball will also have a momentum. To find the momentum of the target ball, we can use the equation:

m1v1 + m2v2 = m1v1' + m2v2'

Where m1 and m2 are the masses of the cue ball and target ball respectively, v1 and v2 are their initial velocities, and v1' and v2' are their final velocities. Since we know the masses and initial velocities of both balls, we can solve for the final velocity of the target ball.

However, we also need to take into account the angle at which the target ball will be moving after the collision. This can be found using the law of conservation of energy, which states that the total energy before a collision is equal to the total energy after the collision. This means that the kinetic energy of the cue ball before the collision is equal to the kinetic energy of both balls after the collision.

To find the angle of the target ball after the collision, we can use the equation:

tanθ = (m1v1' sinθ - m2v2' sinθ) / (m1v1' cosθ + m2v2' cosθ)

Where θ is the angle
 
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