Question

A block of mass m is hanging freely from a spring of stiffness k.A particle of mass m falls on the block with velocity

Answer 1

Answer:

weightlessness block.with velocity

Answer 2

Thus the amplitude of the particle is A = Vo √  m / 2 k and angular frequency is ω =   √  k / 2 m v(max)

Explanation:

Complete statement:

A block of mass m is hanging freely from a spring of stiffness k. A particle of mass m falls on the block with a velocity v and gets stuck to it. The amplitude and angular frequency of oscillations during subsequent motion are respectively.  

Solution:

By conservation of momentum:

mVo = (m + m ) V / V = Vo / 2 ( Velocity after collision)

ω = √  k / m + m =

ω =   √  k / 2 m v(max)

Aω = Vo / 2

A = Vo √  m / 2 k

Thus the amplitude of the particle is A = Vo √  m / 2 k and angular frequency is ω =   √  k / 2 m v(max)