university 2 physics question?

Question

A 5.0g bullet is fired horizontally into a 0.5kg block of wood resting on a frictionless table. The block, which is attached to a horizontal spring, retains the bullet and moves forward, compressing the spring. The block-spring system goes into SHM with a frequency of 9.0Hz and an amplitude of 15cm. Determine the initial speed of the bullet.

Answer 1

Oh, very good!  This one could be done several ways.

The long way is to use the frequency and mass to calculate the spring constant K, determine the energy of the system from the amplitude, and then use that to determine the initial speed of the block.  From that speed, the momentum of the block post-impact (and thus the bullet pre-impact) is derived.  Last, the bullet's velocity is calculated.

The short way is to go straight from the equation of motion (which is implied) to the velocity of the block.  The frequency and amplitude are given and all we have to do is add the boundary condition x=0 at t=0, implying

x(t) = 0.15 sin(18πt)

The velocity is the derivative, or

dx/dt = 2.7π cos(18πt)

which equals 2.7π m/sec at t=0.  The momentum of the block immediately post-impact is 2.7π*0.505 or 4.2836 kg-m/sec.  Dividing this by the mass of the bullet yields 856.7 m/sec.

Answer 2

f=sqrt(k/m)/(2pi)

therefore k=(2pif)^2*m

k=1614.9

The initial velocity of the bullet block system is found by equating energy

1/2kx^2=1/2mv^2

1/2(1614.9)(.15)^2=1/2(.505)V^2

so v=8.48 m/s

The initial velocity is found by considering conservation priciples

v(bullet)m(bullet)=v(system)*(m(bullet)+m(block))

v(bullet)=8.48*(.505)/.005=856.7 m/s

Answer 3

LOL

V. = 0 m/s in the barrel