An object of mass m 1 = 9.10 kg is in equilibrium while connected to a light spr
ID: 1402666 • Letter: A
Question
An object of mass m1 = 9.10 kg is in equilibrium while connected to a light spring of constant k = 100 N/m that is fastened to a wall as shown in Figure P15.52a. A second object, m2 = 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.190 m, (see Fig. P15.52b). 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. P15.52c) and moves to the right with speed v. Determine the value of v.
m/s
(b) How far apart are the objects when the spring is fully stretched for the first time (D in Fig. P15.52d)? (Suggestion: First determine the period of oscillation and the amplitude of the m1-spring system after m2 loses contact with m1.)
cm
Explanation / Answer
When the spring and masses return to the equilibrium point the masses will have KE of
1/2*(m1+m2)*v^2
and the spring will have lost PE of
1/2*k*x^2
1/2*100*0.19^2
set up the equation
1/2*(9.1+7)*v^2= 1/2*100*0.190^2
v = 0.474 m/s
part b )
the period will determine when the spring reaches max extension. Then compute the distance that the m2 has traveled over a period of 1/4 of the total oscillation for this part
w = sqrt(k/m)
the first time the spring reaches full extension after m2 is launched is
w*t=2*pi/4
sqrt(k/m)*t=pi/2
solve for t
t=sqrt(m/k)*pi/2
t = 0.630 s
then the amplitude is truncated from the initial release A because of the loss of m2
using energy conservation
1/2*m1*v^2=.5*k*x^2
solve for x
x = v* sqrt(m1/k)
that is the distance m1 will travel to the full extension of the spring
x= 0.143 m
now compute the distance that m2 has traveled
x2= v * t *(pi/2)
x2 = 0.4688
the distance they are apart is
x2-x
or
d=0.4688 - 0.143
0.3258 m
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