A block of mass M sits at equilibrium between two springs of spring constant k₁ and k₂. A bullet of mass m and speed v is fired horizontally at the block and becomes embedded in it. Thereq is no friction between the block and the surface.

(a) What is the amplitude of the resulting oscillation?

(b) What is the angular frequency?

How should we connect the motion of the bullet before the collision to the motion of the block + bullet after the collision?

1. Conservation of linear momentum
2. Conservation of mechanical energy
3. Either one works

Use momentum conservation, because mechanical energy is lost in the inelastic collision.

Momentum before the collision = momentum after the collision gives:

mv + 0 = (M + m) vf →  vf

=

m M + m

v

The block plus bullet has velocity v_f at the equilibrium position, where the potential energy is zero. We now use energy conservation to find the maximum amplitude: the kinetic energy at the equilibrium position equals the potential energy stored in the two springs at the maximum amplitude point.

K_i= U_f  →  ½ (m+M)v_f² = ½ k₁A² + ½ k₂A²

Cancelling factors of ½ and substituting v_f from above gives:

m2 v2 M + m

=

(k1 + k2) A2

This gives an amplitude of

A

=

m v [(M + m)(k1 + k2)]½

(b) What is the angular frequency?

There are two ways to solve for ω. One is to remember that for simple harmonic motion v_max = Aω.

Since v_max = v_f, the speed at the equilibrium position, we have:

ω

=

vf A

=

m v M + m

*

[(M+m)(k1 + k2)]½ m v

ω

=

(

k1 + k2 M + m

)

½

The other solution method is to analyze the forces due to each spring. Applying Newton's Second Law gives:

ΣF = ma  →  -k₁x - k₂x = (M + m) a

a

= -

k1 + k2 M + m

x

For and equation in this form, the angular frequency is the square root of the factor multiplying - x, so:

ω

=

(

k1 + k2 M + m

)

½