Two masses attached by inextensible string?

[cupcaked]

Homework Statement

okay this is probably extremely basic stuff but I'm having a lot of trouble in this class and i hope someone can help! :/

A 10 kg mass is placed on a fricitonless surface and is attached to a 5 kg mass as shown by a light and inextensible string.
( here is a drawing of the model: http://i53.tinypic.com/ekne6e.png )

a) Find the acceleration with which the system moves.
b) What is the speed of the masses when the 5 kg mass strikes the floor?
c) After the 5 kg mass strikes the floor, the 10 kg mass on the table continues to move to the right. How far from the edge of the table does it land?
d) How would your answer to part c change if there was friction between the mass and the table? Explain.

Homework Equations

The Attempt at a Solution

a) I set up the equations:
T = 10a
49-T = 5a

49-10a = 5a
so i got a = 3.266 m/s^2

b) I used the formula Vf^2 = Vi^2+2ad but I've got a feeling I'm approaching this problem totally wrong. But anyway:
Vf^2 = (3.266^2)+(2*9.8*0.6)
Vf^2 = 22.43
Vf = 4.74 m/s^2 (i put that as my answer)

c) I have no clue how to approach this, if anyone can guide me please? All I can think of is using the same equation as the last part but I don't think I'm right?

d) I put: If the surface had friction, the answer to part (c) would be less than the answer obtained in part (c), because the friction would be an opposing force against where the mass is trying to move.
Not the best explanation but it answers the question, I think.

if anyone can help I would appreciate it SO much! I'm currently failing ap physics and I really don't want to be :'(

 

[Delphi51]

(a) looks good
(b) "Vf^2 = Vi^2+2ad
Vf^2 = (3.266^2)+(2*9.8*0.6) "
Since Vi = 0, you should have Vf^2 = 2*3.266*0.6

(c) The 10 kg mass will no longer accelerate because the 5 is no longer pulling on it. So you know its initial velocity when it flies off the table. I trust it doesn't hit the pulley. This leaves you with a 2D trajectory problem to solve. My approach for them is to write two headings for horizontal and vertical, ask yourself what kind of motion you have in each and write the formulas for that motion in each case. Put in the knowns in all three formulas and hope one of them can be solved for something.

Good luck!

 

[cupcaked]

Yay! Okay so

b)
Vf^2 = 0+(2*3.266*0.6)
Vf^2 = 3.91
Vf = 1.97 m/s^2

And for part (c)... I separated the x component and y compenent of the motion into separate charts, though I completely failed the 2d motion test so if my work can be checked/corrected, I'd appreciate it! here's my work: http://i55.tinypic.com/250k1fk.png

 

[Delphi51]

I got 1.98 in place of your 1.97, but I use g = 9.81 so we agree.

Oops, in the horizontal calculation you used an acceleration of 9.81!
Gravity only acts in the vertical direction. Horizontally, just use x = v*t.
I agree with your time to 2 digit accuracy. Just multiply it by 1.97.

Sorry to hear you had trouble with 2D motion. I am a veteran of 29 years teaching high school physics, and my students had really good success with it once I got them into that routine beginning with the 2 headings and asking what type of motion in each direction. Usually it is accelerated motion vertically and constant speed motion horizontally.

 

[asteriatic]

Hi there, I can definitely help you with this problem. First, let's talk about what is meant by an inextensible string. This means that the string has no stretch or flexibility, so the distance between the two masses remains constant. Now, let's go through each part of the problem.

a) To find the acceleration, we can use Newton's second law: F=ma. In this case, the only force acting on the system is the tension in the string. So we can set up the equation: T=ma. Since the masses are connected by the string, they will have the same acceleration. So we can set the two equations equal to each other: T=10a and T=5a. Now, we can solve for the acceleration by setting these two equations equal to each other: 10a=5a. This gives us a=3.266 m/s^2, which is the correct answer.

b) Your approach to this part is correct. To find the final velocity, we can use the equation Vf^2 = Vi^2+2ad. However, we need to be careful with the values we plug in. The initial velocity (Vi) is 0 since the masses start from rest. The distance (d) is the distance the 5 kg mass moves before hitting the floor, which is 0.6 m. So the correct equation should be Vf^2 = 0^2+2*3.266*0.6. This gives us a final velocity of 2.4 m/s, which is the correct answer.

c) To find the distance the 10 kg mass will travel before hitting the floor, we can use the equation d=Vit+1/2at^2. Again, the initial velocity is 0 and the acceleration is 3.266 m/s^2. We know that the final velocity when the 5 kg mass hits the floor is 2.4 m/s, so we can use this to solve for the time (t): 2.4=3.266t. This gives us t=0.734 seconds. Now, we can plug this into the original equation to find the distance: d=0+1/2*3.266*(0.734)^2. This gives us a distance of 1.5 m, which is the correct answer.

d) Your explanation for this