## saljudieh07 2 years ago SPRING TWO MASSES AND A COLLISION, I am gonna go nuts if not one helps me! xD

1. saljudieh07

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)?

2. saljudieh07

@Mertsj @hartnn @UnkleRhaukus @Callisto @robtobey

3. saljudieh07

4. Mertsj

I would help you if I could but I am terrible at physics.

5. Mertsj

If you type the problem into a google search, you will probably find its solution somewhere.

6. NotTim

7. stgreen

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 one-fourth 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

8. stgreen

^courtesy quicksilver

9. saljudieh07

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.

10. saljudieh07

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?

11. stgreen

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..

12. saljudieh07

ohh, i see... so it is because of inertia that m2 keeps moving with the same speed as both of m1 and m2

13. stgreen

^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

14. saljudieh07

yes it makes sense, so at x=0

15. saljudieh07

also, m2 had to be at a higher speed coming to hit m1, right?

16. stgreen

during compression?yes initially

17. saljudieh07

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

18. saljudieh07

is my analysis right?

19. stgreen

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)

20. saljudieh07

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

21. saljudieh07

which I did and found vi to be 0.9063 m/s

22. saljudieh07

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?

23. saljudieh07

Thanks a lot too =D

24. stgreen

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

25. saljudieh07

hahaha, well it works for you. i think i need to go over my physics laws because i confuse them sometimes.

26. stgreen

lol

27. saljudieh07

ok thanks again, see you later

28. stgreen

see ya