Question

Questions 13-16 A massiess spring with spring constant k is attached at one end of a block of mass M that is at rest on a frictionless horizontal table. The other end of the spring is fixed to a wall. A bullet of mass ma is fired into the block from the left with a speed to and comes to rest in the block M 7m 14. Find the amplitude of the resulting simple harmonic motion. to (c mato (d) st (e

Answer 1

Conservation of momentum,

Initial momentum = mb*v₀

Final momentum = (mb+M)v

mb*v₀ = (mb+M)v
Therefore immediately after the collision, the bullet and the block move at a speed of v = mb*v0/(mb+M)

The kinetic energy thus produced is converted to spring potential energy,

The amplitude of the spring is when the compression in the spring is maximum and the kinetic energy becomes 0.

1/2 (mb+M) v² = 1/2 k A²

mb2 v₀²/(mb+M) = kA²

A = sqrt(1/k(mb+M)) mbv₀

Option c is correct.