Question

9. A mass m is attached to a massless spring with a force constant k. The mass rests on a horizontal, frictionless surface. The system is compressed a distance x from the spring's initial position and then released. The momentum of the mass when the spring passes its equilibrium position is given by (A) xvmek (B) x/k/m o x/m/k (D) x/km + KxP = {mv² p=mv

Answer 1

* Solution : - к 10 om Im (initial position) Doponge Push C compressed Position) x 2 (fig. I) The potential energy stored in the spring after compression and it is given by - U= kuc where, K = Spring consternt. x = displacement of the block after compression when Spring passes its equilibrium position the potential energy converted into Kinetic energy of the block and it is given by - k = $mumas) where, m = mass of the blocki Umax = maximum velocity of the block when it reaches its equilibriem position. Now, from conservation of energy energy U=k 2

them 2 1 kre? 12 m (Umax)? Umax - K c2 m. Now, we com calculate the momentum of the mass as P= m(Umax). Р m Kr2. m P m² kac² m P = =In Im m Im kr? P = x mk. Ans. Ang - Option (A) 3c Imk. is correct.