An object of mass m 1 = 8.40 kg is in equilibrium while connected to a light spr

Question

An object of mass m1 = 8.40 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 = 0.240 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

Apply law of conservation of energy

Total energy of the system when it was strecthed = Total energy of the system when it was in equilibrium

(0.5*k*A^2) = 0.5*(m1+m2)*v^2

0.5*100*0.24^2 = 0.5*(8.4+7)*v^2

2.88 = 7.7*v^2

v = sqrt(2.88/7.7) = 0.611 m/s

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Time period of m1 is T = 2*pi*sqrt(m1/k) =2*3.142*sqrt(8.4/100) = 1.82 S

After T/4 = 1.82/4 = 0.455 S later the mass m1 passes the point whre the spring is fully streched first time

at the equilibrium

the total energy is 0.5*m1*v^2 = 0.5*8.4*0.611^2 = 1.567 J

but this is equal to energy at the fully streched point

1.567 = 0.5*k*A'^2

1.567 = 0.5*100*A'^2

A' = sqrt(1.567/50) = 0.177 m

then required distnce is v(T/4) - A' = (0.611*0.455)-0.177 = 0.101 m = 10.1 cm

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