An object of mass m 1 = 9.10 kg is in equilibrium when connected to a light spri

Question

An object of mass m1 = 9.10 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 part [a] of the figure below. A second object, m2 = 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.140 m (see part [b] of the figure below). The system is then released and both objects start moving to the right on the frictionless surface.

An object of mass m1 = 9.10 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 part [a] of the figure below. A second object, m2 = 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.140 m (see part [b] of the figure below). The system is then released and both objects start moving to the right on the frictionless surface.both objects start moving to the right on the frictionless surface. When m1 reaches the equilibrium point, m2 loses contact with m1 (see part [c] in the above figure) and moves to the right with speed v. Determine the value of v. How far apart are the objects when the spring is fully stretched for the first time (the distance D in part (d) of the above figure)?

Explanation / Answer

a) By conservation of energy at the equilibrium point

1/2*kx^2 = 1/2*(m1+m2)v^2

100*0.14^2 = 16.1*v^2

v = 0.35 m/s

b) The time taken by body 1 to reach extreme end is given by 1/4th of time period

w = sqrt(k/m) = 3.315

t = pi/(2*w) = pi/(2*3.315) = 0.474 s

The distance travelled by body 1 from equilibrium be x

1/2*100*x^2 = 1/2*9.1*0.35^2

x = 0.1056 m

Distance moved by body 2 = v*t = 0.35*0.474 = 0.1659

D = 0.1659 - 0.1056 = 0.0603 m = 6.03 cm

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