Two objects of masses m1 = 0.48 kg and m2 = 0.92 kg are placed on a horizontal f

Question

Two objects of masses m1 = 0.48 kg and m2 = 0.92 kg are placed on a horizontal frictionless surface and a compressed spring of force constant k = 290 N/m is placed between them as in figure (a) shown below. Neglect the mass of the spring. The spring is not attached to either object and is compressed a distance of 9.0 cm. If the objects are released from rest, find the final velocity of each object as shown in figure (b). (Let the positive direction be to the right. Indicate the direction with the sign of your answer.)
v1 = m/s
v2 = m/s

Explanation / Answer

conserve the linear momentum of the system.
pf=pi
=>m1v1 + (-m2v2)= 0 + 0
=>0.48v1 = 0.92v2
=>v1=1.91v2

now apply energy conservation.
1/2 kx2 = 1/2 m1v12 + 1/2 m2v22
=> kx2 = 3.67m1v22 + m2v22
=> kx2 = (3.67m1+m2)v22
=> v2 = -9.35 m/s (it is to the left side)
=> v1 = 17.87 m/s (it is to the right side).

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