Two objects of masses m1 = 0.48 kg and m2 = 0.92 kg are placed on a horizontal f
ID: 2058791 • Letter: T
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 9.4 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=
v2 =
Explanation / Answer
energy conservation gives :
(1/2)Kx2 = (1/2)m1v21 + (1/2)m2v22
and from momentum conservation :
0 = m1v1 - m2v2
solving the above equations give :
v1 = 1.87 m/s
v2 = 0.98 m/s