Two spring systems are configuredsuch that a ball is tossed back and forth betwe

Question

Two spring systems are configuredsuch that a ball is tossed back and forth between each springsystem. No energy losses exist in the mechanisms. The springssystems are angled 60o degreeswith respect to the horizontal axis. The springconstant is 100N/m. The ball has a mass of 2kg and travels ahorizontal distance of 2m between the spring systems. When the ballstrikes the spring, the velocity vector is along the axis of thespring, 60o degrees relative to the horizontal surface.

Find thefollowing:

1) Maximum velocity of the ball

2) Maximum height of the ball

3) Maximum deflection of a spring

4) Total energy in the system

Explanation / Answer

ok bear with me on this my projectile motion is a littlerusty.... so the problem you have uses kinimatics and the engergyeq's the Potential of a spring is 1/2*k*s2, the force ona spring is k*s,                            grav     mgh                             grav      9.81 m/s2(ay but remember negitive) so our kinimatics eq's are: x = voxt y = v0yt +1/2*ayt2 vx = vox (for constant accel inx-direction) vy = voy + ayt so now using these eq's we can get the height thenspring deflection then engergy (the max velocity eludes me rightnow but its going to be at impact) so since we kno at max height vy is 0then lets get 't' in terms of values we kno using...       vy =voy + ayt => 0= voy +ayt => t = voy/-ay  = vosin60 /g now we can use y = v0yt +ayt2to get the max height y = v0y(vosin /g)+1/2*ay(vosin /g)2 = -(vosin)2/2g +(vosin /g )2 =(vosin)2/2g sin = 1 when max height soymax (hmax)= vo2/2g for the max displacment use the engergy eq, since we know thatther are no losses the engergy going in is the same going outso... Ein = UE + KE = (1/2*k*s2 + mgh)+0; Eout = 0 + 1/2*m*v2   (or visaversa depends on how you view it) then solve for 's' so for the total engergy you would either use Einor Eout the max velocity eludes me at the moment, but i hope this getsyou on track to fiqure out your problem.... for the max displacment use the engergy eq, since we know thatther are no losses the engergy going in is the same going outso... Ein = UE + KE = (1/2*k*s2 + mgh)+0; Eout = 0 + 1/2*m*v2   (or visaversa depends on how you view it) then solve for 's' so for the total engergy you would either use Einor Eout the max velocity eludes me at the moment, but i hope this getsyou on track to fiqure out your problem....

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