Two objects of masses m 1 = 0.52 kg and m 2 = 0.90 kg are placed on a horizontal

Question

Two objects of massesm1=0.52kg andm2=0.90kg are placed on a horizontal frictionless surface and a compressed spring of force constantk=300N/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 of9.2cm. 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

since the force of spring is internal force which means that external force on the objects is zero so momentum will be conserved. initial velocity of both object is zero final velocity of m1 and m2 is v1(left) and v2(right) respectively so by conservation of momentum 0 = m1v1 -m2v2 m1v1= m2v2. v1= (m2/m1)v2 v1 = (45/26)v2 also potential energy of compressed spring will also be converted into kinetic energy of objects. so net potential energy = net kinetic energy 1/2kx^2 = 1/2(m1v1^2 + m2v2^2) 300*(9.2/100)^2 = 0.52(45v2/26)^2 + 0.90v2^2 2.5392 = 2.4576v2^2 v2 = 1.01m/s v1 = 45v2/26 = 1.748m/s. = -1.748m/s.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.