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, becoming embedded in it. Assume there 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 (15/31) (48%)
2. Conservation of mechanical energy (8/31) (26%)
3. Either one would work (8/31) (26%)

We have to use momentum conservation here, because mechanical energy is lost in the completely inelastic collision.

Momentum before the collision = momentum after the collision.

mv + 0 = (M + m) v_f

vf

=

m M + m

v

The block plus bullet has this velocity at the equilibrium position, where the potential energy is zero. We can now use conservation of energy to find the amplitude - the kinetic energy at the equilibrium position equals the potential energy stored in the two springs at the end-point of the motion, where the mass comes instantaneously to rest.

K_i = U_f

½ (m+M)v_f² = ½ k₁A² + ½ k₂A²

Cancelling factors of ½ and substituting our expression for 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 approach this. One is simply to remember that for simple harmonic motion v_max = Aω.

v_max = v_f, the speed at the equilibrium position, so:

ω

=

vf A

=

m v M + m

*

[(M+m)(k1 + k2)]½ m v

ω

=

(

k1 + k2 M + m

)

½

The other way to approach this is to analyze the forces. Each spring applies a restoring force to the system, so applying Newton's Second Law gives:

ΣF = ma

-k₁x - k₂x = (M + m) a

a

= -

k1 + k2 M + m

x

When we get the equation in this form the angular frequency is the square root of whatever is multiplying -x, so:

ω

=

(

k1 + k2 M + m

)

½