An object with mass m1 = 9.00 kg is in equilibrium while connected to a horizontal light spring of constant k = 100 N/m that is fastened to a wall. A second object, of mass m2 = 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.200 m. The system is then re- leased, and both objects start moving to the right on the frictionless surface. When mị reaches the equilibrium point, m2 loses contact with m1 and moves to the right with speed v. (a) Determine the value of v. (b) How far apart are the objects when the spring is fully stretched for the first time.

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Chapter39: Relativity
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Not sure how to approach part b.

An object with mass m1 = 9.00 kg is in equilibrium while connected to a horizontal light spring of
constant k
100 N/m that is fastened to a wall. A second object, of mass m2 = 7.00 kg, is slowly
pushed up against m1, compressing the spring by the amount A = 0.200 m. The system is then re-
leased, and both objects start moving to the right on the frictionless surface. When m¡ reaches the
equilibrium point, m2 loses contact with mį and moves to the right with speed v.
(a) Determine the value of v.
(b) How far apart are the objects when the spring is fully stretched for the first time.
Transcribed Image Text:An object with mass m1 = 9.00 kg is in equilibrium while connected to a horizontal light spring of constant k 100 N/m that is fastened to a wall. A second object, of mass m2 = 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.200 m. The system is then re- leased, and both objects start moving to the right on the frictionless surface. When m¡ reaches the equilibrium point, m2 loses contact with mį and moves to the right with speed v. (a) Determine the value of v. (b) How far apart are the objects when the spring is fully stretched for the first time.
Expert Solution
Step 1

(a)

When the second object pushed against first object, the work done is stored as potential energy in the spring.

The stored potential energy is converted in to kinetic energy when the released.

Apply principle of conservation of energy

12m1+m2v2=12kx2Rearrangev=km1+m2x2

Substitute

v=100 Nm9.00 kg+7.00 kg0.200 m2=0.5 ms

This is the speed of both the masses.

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