Two objects of masses m 1 = 0.60 kg and m 2 = 0.94 kg are placed on a horizontal

Question

Two objects of masses m1 = 0.60 kg and m2 = 0.94 kg are placed on a horizontal frictionless surface and a compressed spring of force constant k = 250 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 9.2 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.)

Explanation / Answer

apply from the formula of law of conservation of energy EPE = KE = PE

0.5 kx^2 = (0.5 m1u1^2 + 0.5 m2u2^2)

u1^2 *(0.6/2) + u2^2 (0.94/2) = 0.5* 250 * 0.092* 0.092

0.3 u^2 + 0.47 v^2 = 1.058 -----------------------1

now from the conservation of momentum

m1u + m2v = 0

0.6 u + 0.94 v = 0

V = (-0.64 u)

so by substituing in 1 we get

0.3 u^2 + (0.47 * 0.64* 0.64* u^2) = 1.058

u = 1.465 m/s ------------<<<<<<<<<<<<<Answer

so

v = -0.64 * 1.465 = -0.9376 m/s-------------<<<<<<<<<<<<<Answer

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