Understanding Physics for the Beer Mug Problem

AI Thread Summary
A bartender slides a beer mug at 1.50 m/s off a frictionless bar that is 1.2 m tall, prompting a discussion on calculating the mug's speed and direction upon impact. The equations for horizontal and vertical velocity components are provided, but additional equations are needed to determine Vx and Vy at the moment of impact. The correct impact speed is identified as 5.08 m/s with a direction of θ = -72.8°. Participants are encouraged to consider the relationship between the sliding velocity and Vx without immediately resorting to equations. The discussion emphasizes understanding the physics behind projectile motion in a frictionless environment.
Let It Be
Messages
45
Reaction score
0
1. A bartender slides a beer mug at 1.50m/s towards a customer at the end of a frictionless bar that is 1.2m tall. the customer makes a grab for the mug and misses, and the mug sails of the end of the bar. What are the speed and direction of the mug at impact?


2. Vx=Vocosθ
Vy=Vosinθ
V=√Vx^2+Vy^2
θ=tan-1(Vy/Vx)


3. The correct answer apparently is 5.08m/s and θ=-72.8°. I didn't get that...
 
Physics news on Phys.org
The equations you give are OK to find the speed and direction, but you need some more equations to find Vx and Vy when the mug hits the floor.
 
AlephZero said:
The equations you give are OK to find the speed and direction, but you need some more equations to find Vx and Vy when the mug hits the floor.

I don't know how to solve the problem with the equations I do have though...
 
How do you think vx and the sliding velocity compare?

Think about it, don't try use the equations!
 
217 MeV said:
How do you think vx and the sliding velocity compare?

Think about it, don't try use the equations!

I know I need to find Vf for the speed when the mug hits the floor. And then direction idk
 
Well, start with the x-component of v_f. What will that be?

Remember there's no friction whilst the mug is sliding, and I assume there is no air resistance.
 
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