Question

A bullet with a mass mb=13.5 g is fired into a block of wood at velocity vb=245 m/s. The block is attached to a spring that has a spring constant k of 205 N/m. The block and bullet continue to move, compressing the spring by 35.0 cm before the whole system momentarily comes to a stop. Assuming that the surface on which the block is resting is frictionless, determine the mass mw of the wooden block.

Answer 1

Momentum is conserved, so the sum of the momenta of the bullet and block before collision is equal to the momentum of the combined bullet-block system,

$m_bv_b+m_wv_w = (m_b+m_w)v$

where v is the speed of the bullet-block system. The block starts at rest so it has no initial momentum, and solving for v gives

$v = \dfrac{m_b}{m_b+m_w} v_b$

The total work W performed by the spring on the bullet-block system as it is compressed a distance x is

$W = -\dfrac12kx^2$

where k is the spring constant, and the work done is negative because the restoring force of the spring opposes the bullet-block as it compresses the spring.

By the work-energy theorem, the total work done is equal to the change in the bullet-block's kinetic energy ∆K, so we have

$W_{\rm total} = W = \Delta K$

The bullet-block starts moving with velocity v found earlier and comes to a stop as the spring slows it down, so we have

$-\dfrac12kx^2 = -\dfrac12(m_b+m_w)v^2 \implies kx^2 = \dfrac{{m_b}^2}{m_b+m_w}{v_b}^2$

Solve for $m_w$:

$m_w=\dfrac1k\left(\dfrac{m_bv_b}x\right)^2-m_b$

$m_w=\dfrac1{205\frac{\rm N}{\rm m}}\left(\dfrac{(0.0135\,\mathrm{kg})\left(245\frac{\rm m}{\rm s}\right)}{0.350\,\rm m}\right)^2-0.0135\,\mathrm{kg}\approx \boxed{0.422\,\mathrm{kg}}$