A block of mass 0.2 kg is attached to a spring of spring constant 23 N/m on a fr

Question

A block of mass 0.2 kg is attached to a spring of spring constant 23 N/m on a frictionless track. The block moves in simple harmonic motion with amplitude 0.16 m. While passing through the equilibrium point from left to right, the block is struck by a bullet, which stops inside the block. The velocity of the bullet immediately before it strikes the block is 68 m/s and the mass of the bullet is 3.35 g.

a)find the speed of the block immediately before the collision.

b)If the simple harmonic motion after the collision is described by x=Bsin(omega*t+phi), what is the new amplitude B?

c)THe collision occurred at the equilibrium position. How long will it take for the block to reach maximum amplitude after the collision?

Explanation / Answer

conserving energy,

0.5mv^2=0.5kx^2

or 0.5*0.2*v^2=0.5*23*0.16^2

or v=1.71 m/s

now conserving momentum,

0.71*0.2+68*3.35*10^-3=v*(0.2+3.35*10^-3)

or v=1.82 m/s

b)again conserving energy,

0.5*(0.2+3.35*10^-3)*1.82^2=0.5*23*x^2

or x=0.1711 m

c)T=0.5*2pi*(m/k)^0.5

=0.147 s

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