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Spring mechanics with 2 different masses

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1. Homework Statement
Two masses m and 3m are joined together by a compressed massless spring of constant k = 750 N/m. When spring is released it pushes masses apart and they slide on a horizontal tabletop as shown below. Mass 3m slides to the right and its track is frictionless. Mass m slides to the left and part of its track of length of .75 m has a friction where mu = .15. Compression of the spring is .25 m. For the mass 3m there is a ramp at the end of the tabletop. The ramp makes an angle of 20 degrees with the horizontal. Mass m is .5 kg. A length of the table is 2 m. Spring regains its initial length before mass m starts to slide with friction and mass 3m reaches the ramp. Using the data given above answer the followings:
1. calculate the potential energy of the spring.
2. Calculate the speed of the mass m before it starts to slide with friction.
3. Calculate speed of the mass 3m before it enters the ramp.
4. Calculate the work of the friction force for the mass m.
5. Calculate the speed of the mass m when it leaves the table.
6. Calculate the speed of the mass 3m when it leaves the ramp.
7. For the mass m find the time that is required to reach the floor (from the edge of the tabletop to the floor).
2. Homework Equations and attempt at a solution
for problem 1: PE =KE, where PE = potential energy and KE = kinetic energy formula two equals .5K delta X^2. this means PE = (750*.25^2)/2 = 23.44 J

for problem 2 I believe you use V = square root (2KE)/m = square root 2(23.44)/.5 = 9.68m/s

I believe problem 3 is solved the same way but that we change the mass so V = 5.59 m/s

problem 4 I believe would use 2 formulas W = fd and F(f) = mu* mg, where F(f) = force friction so it would be mu *mg *d = .552 J

problem 5 if I am not mistaken would use the equation V = (square root 2g delta h) = square root 2 *9.81 * .9 = 4.2

Problem 6 is where I'm a little hazy. Also I am not sure If I'm using the right equations. I have not done mechanics since November so if some one could check my work I would very much appreciate it. Also please excuse the English my teacher is Russian and I wrote the problem word for word.
 

Answers and Replies

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this picture should help
 

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Is anyone there?
 
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I believe I made a mistake on # 5 so what I think I was suppose to do was take the work of the spring - the work of friction = (mv^2)/2 now multiply the new work by 2 divide by m and square root both sides and you get velocity. so the answer should be 9.568 round to 9.57.
 
Astronuc
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The energy of the spring becomes the kinetic energy of both mass, m and 3m. Conservation of energy. The same force is applied to both block, F = kx. Also, the total (net) momentum before the blocks move is zero, and then think about conservation of momentum.
 
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I'm confused how does that help me find the speed of mass 3m when it leaves the ramp? Your answer seems to be the perspective I need to look at this problem, is their something wrong with how I have been solving the problems? If so where did I mess up and if not using work/energy formulas how do I find the velocity of mass 3m going off the 20 degree ramp?
 
Kurdt
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Take this one step at a time. What is the potential energy of a compressed spring?
 
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I've already solved that, I'm on step 6 but the answer is 22.44, all the info you need is in the first, second, and fourth post.
 
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The way I see it I found the x direction velocity in problem 3 so I need to find V that accommodates two dimensions (the x as well as the y). so i am using the formula v0cos 20 = vx so vx/cos20 = v0 which gives me 5.55m/s. So I'm wondering if my method is correct?
 

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so T should = .43 because y=.5at^2, so t = square root y/(.5a)
 
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final answers:
for problem 1: PE =KE, where PE = potential energy and KE = kinetic energy formula two equals .5K delta X^2. this means PE = (750*.25^2)/2 = 23.44 J

for problem 2 I believe you use V = square root (2KE)/m = square root 2(23.44)/.5 = 9.68m/s

I believe problem 3 is solved the same way but that we change the mass so V = 5.59 m/s

problem 4 I believe would use 2 formulas W = fd and F(f) = mu* mg, where F(f) = force friction so it would be mu *mg *d = .552 J

problem 5 take the work of the spring - the work of friction = (mv^2)/2 now multiply the new work by 2 divide by m and square root both sides and you get velocity. so the answer should be 9.568 round to 9.57.

problem 6 I am using the formula v0cos 20 = vx so vx/cos20 = v0 which gives me 5.55m/s.

problem 7 T should = .43 because y=.5at^2, so t = square root y/(.5a)

I just want someone to tell me if 6 and 7 are correct
 
Kurdt
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For parts two and three you haven't applied the conservation of energy and momentum as Astronuc suggested.
 
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why doesn't my answer take into account the conservation of energy and momentum/how do It the right way?
 
Kurdt
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why doesn't my answer take into account the conservation of energy and momentum/how do It the right way?
The kinetic energy of the two masses must add up to the potential stored in the spring. You have calculated the kinetic energy for each mass as if they both gain all the potential energy.
 
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OK I'm not quite sure how to do that, But this is my attempt. Because of conservation of momentum I know PE = (.5V1^2)/2 + (1.5V2^2)/2 but that leaves me with two variables, so what do I do? M =.5 and 3M = 1.5, which means I am using the formula KE = (mv^2)/2
 
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Kurdt
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one must also conserve momentum.
 
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Can you give me a formula or a procedure that I can use, cause I know I'm not doing this right, but I want to know the right way.
 
Kurdt
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I'm sorry I can't figure this out.
 
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I'm running out of time can't you just tell me what I need to do? I can do the rest and I have shown an attempt at the problem. Also I know that 1,4,5,and 7 are correct. and their were actually 11 problems and I can solve those If I can find the answer to 2 and 3.
 
tiny-tim
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Two masses m and 3m are joined together by a compressed massless spring of constant k = 750 N/m.
When spring is released it pushes masses apart and they slide on a horizontal tabletop as shown below. …
Compression of the spring is .25 m. …
Mass m is .5 kg.
Spring regains its initial length before mass m starts to slide with friction and mass 3m reaches the ramp. …
2. Calculate the speed of the mass m before it starts to slide with friction.
3. Calculate speed of the mass 3m before it enters the ramp.
I'm running out of time can't you just tell me what I need to do? I can do the rest and I have shown an attempt at the problem. Also I know that 1,4,5,and 7 are correct. and their were actually 11 problems and I can solve those If I can find the answer to 2 and 3.
Hi fonzi! :smile:

(Thanks for the PM.)

As Kurdt and Astronuc advised, you need to apply both conservation of energy and conservation of momentum.

(That is true in nearly all collision problems!)
The kinetic energy of the two masses must add up to the potential stored in the spring.
one must also conserve momentum.
So, from energy you get PE = KE1 + KE2.

And from momentum you get m1v1 = -m2v2.

The momentum equation is linear, while the energy equation is quadratic, so solve the monentum equation first: v1 = -(m2/m1)v2.

Then plug that into the energy equation! :smile:

hmm … you seem fine with the other stuff … but you're obviously unhappy about this procedure … I don't understand why … :confused:
 
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thanks tiny tim

problem 1 is the same as before

for problem two and three we use the formula PE = (mV1^2)/2 + (3mV2^2)/2 to calculate conservation of energy and mv1 =3mv2 to calculate conservation of momentum. this will give you 8.39 m/s for problem 2 and 2.8 m/s for problem three.

problem 4 is the same

problem 5 is where I might have a problem. I solved this step by taking Work of the spring for mass m with respect to conservation of energy and momentum minus the work of friction = mv^2/2 solve for v. so V = 8.26m/s

problem 6 could also be wrong. I solved this step by KEi +PEi = KEf + PEf. where PEi =0 and KEi = the kinetic energy of mass 3m. ok now KEi = (mv^2)/2 = 1.5 2.8^2.5 =5.88. Now PEf =mgh and this works because the mass is leaving the ramp so you have potential. The height is .3 because thats the change in height from the table to the ramp. mgh = 1.5 *9.81 *.3 =4.41. now subtract PEf from KEi so that your formula reads KEf = KEi- PEf. this means (1.5v^2)/2 = 1.47 now solve for v which will give you 1.4m/s

problem 7 is the same as before

So tell me what you guys think?
 
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Problem 8 is for the Mass 3m find the time that is required to reach the floor (from the top of the ramp to the floor).

I use three formulas v0sin theta =V0y, y = v0yt =1/2 at^2, and the quadratic formula and you solve for t.

Problem 9 is Calculate the velocity of the mass 3m at the point of impact

I use the formula Vf = V0 +at

Problem 10 is Calculate the velocity of the mass m at the point of impact

same formula

problem 11 is find the horizontal distance between the masses when they are on the floor

for this I take the time I found in problem 7 and 8 multiplied by the horizontal velocity of each block to get the distance the two blocks leave from the table then I add two because the table is 2 meters. so the formula looks like Vx1t1 +Vx2t2 +2
 
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DISREGARD THE LAST POST ITS WRONG. this is a solved problem thanks for the help everyone.
 
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