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saljudieh07
Group Title
SPRING TWO MASSES AND A COLLISION, I am gonna go nuts if not one helps me! xD
 one year ago
 one year ago
saljudieh07 Group Title
SPRING TWO MASSES AND A COLLISION, I am gonna go nuts if not one helps me! xD
 one year ago
 one year ago

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saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
An object of mass m1= 9.00 kg is in equilibrium when connected to a light spring of constant k = 100 N/m that is fastened to a wall as shown in Figure P15.73a. A second object, m2= 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.200 m. The question is asking: The system is then released, and both objects start moving to the right on the frictionless surface. (a) When m1 reaches the equilibrium point, m2 loses contact with m1 (see Fig. P15.73c) and moves to the right with speed v. Determine the value of v. (b) How far apart are the objects when the spring is fully stretched for the first time (the distance D in Fig. P15.48d)?
 one year ago

saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
@Mertsj @hartnn @UnkleRhaukus @Callisto @robtobey
 one year ago

saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
Please help guys
 one year ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.0
I would help you if I could but I am terrible at physics.
 one year ago

Mertsj Group TitleBest ResponseYou've already chosen the best response.0
If you type the problem into a google search, you will probably find its solution somewhere.
 one year ago

NotTim Group TitleBest ResponseYou've already chosen the best response.0
or yahoo answers
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.3
We will apply conservation of energy. When the spring is compressed, it will gain some potential energy. When the spring regains its natural length, it will do work on masses to increase their kinetic energy. They will lose contact with each other when the spring comes to its natural length because after getting natural length, the spring will pull the m1 block whereas there is no one to pull m2. Also, their velocities will be equal. kx²/2 = m1v²/2 + m2v²/2 ...(1) 100(0.2)² = 9v² + 7v² 100 x 0.2 x 0.2 = 16v² 4 = 16v² 1/4 = v² v = 0.5 m/s 2) The spring stretches as much as it was compressed. So, the m1 mass will be 0.2 m away from the dotted line. We have to calculate time taken for the spring to stretch fully from natural length. This one is a bit tedious. First, I am going to take onefourth SHM of the spring block system. This will mean the time taken for the block to reach 0.2m distance from dotted line. T = 2π√(m/k) T / 4 = π/2√(m/k) = (3.14 / 2 ) √(9 / 100) = 0.47 seconds In this time, m2 will travel distance = 0.47 x 0.5 = 0.235 m So, D = 0.235  0.2 = 0.035 m
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.3
^courtesy quicksilver
 one year ago

saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
hey stgreen that was great, but i have a couple of questions... Now, why did you assume that the speed of m2 after it leaves m1would be the same as their combined speeds? The way I think about it is like this: you have m2 comes in with v2 hits m1, they move together at a lower speed Vc, then when m2 lets go off m1, it goes back to its normal speed v2, and the spring should stop oscillating exactly when it hits x=0.
 one year ago

saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
I know that my intuition is wrong, but I rely a lot on it when I solve physics problems, I do not know how to not do that, how do you really do physics problems?
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.3
dont forget v, the velocity we calculated is the velocity just after m2 loses contact with m1..and this happens at equilibrium position x=0..now do you agree that m1 and m2 have same v before they reach the point where m1 leaves m2??(since they are in contact,they almost look like a single mass till then)..if you do...m2 will have the same v due to inertia..
 one year ago

saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
ohh, i see... so it is because of inertia that m2 keeps moving with the same speed as both of m1 and m2
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.3
^ops sorry i wrote it wrong in the second line..it should be..m2 loses m1 after m1 reaches the distance equal to the same distance by which itwas compressed
 one year ago

saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
yes it makes sense, so at x=0
 one year ago

saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
also, m2 had to be at a higher speed coming to hit m1, right?
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.3
during compression?yes initially
 one year ago

saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
so the lost speed of m2, which is actually kinetic energy gets absorbed by the spring and then is used to set m1 into oscillation
 one year ago

saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
is my analysis right?
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.3
it sets m1 into oscillation..but m2 moves a distance too...and it wont come back(we wont call it oscillation for m2..though for m1 it is)
 one year ago

saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
so if I wanted to find the initial velocity of the m2 mass, i would do the following: \[0.5m_{2}(0.5)^2 + 0.5kA^2= 0.5m_{2}v_{i}^2\] and solve for vi
 one year ago

saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
which I did and found vi to be 0.9063 m/s
 one year ago

saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
How do you really approach a physics question? do you start by graphing then your intuition? or how? and what if your intuition is wrong? how to avoid getting stuck. in other words, do you have a strategy for solving physics questions?
 one year ago

saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
Thanks a lot too =D
 one year ago

stgreen Group TitleBest ResponseYou've already chosen the best response.3
my pleasure :)...well you should start reading the statement sentence by sentence and then making your own analogy that is easier for you..read a statement and visualize,make a mental image..then apply physics laws and thats it..i dont think it makes any sense :D
 one year ago

saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
hahaha, well it works for you. i think i need to go over my physics laws because i confuse them sometimes.
 one year ago

saljudieh07 Group TitleBest ResponseYou've already chosen the best response.0
ok thanks again, see you later
 one year ago
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