An oscillator of mass M is at rest in its equilibrium position in a potential, V = 1 2 k x – X 2 . A particle of mass m comes from the right with speed u and collides completely inelastic with M and s

Question

An oscillator of mass M is at rest in its equilibrium position in a potential, V = 1 2 k x – X 2 . A particle of mass m comes from the right with speed u and collides completely inelastic with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is: M = 10 , m = 5 , u = 1 , k = 1

TestHub · Accepted Answer

In the first collision mu momentum will be imparted to the system. In the second collision when the momentum of ( M + m ) is in the opposite direction mu momentum of the particle will make its momentum zero. On 13 th collision, Applying momentum conservation m u = ( M + 13 m ) v ⇒ v = m u M + 13 m = u 15 ⇒ M = 2 m ⇒ v = ω A ⇒ u 15 = K M + 13 m × A ⇒ A = 1 15 75 1 = 1 3

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