Problem 8.24

8.24. a)

We are going to assume equal firing speeds and equal forces due to the wood fibers. Considering the case of the block held in the vice,

$K_{i} + \Delta K_{nc} ~=~ K_{f}$

$\frac{1}{2} (.007 kg) v^{2} - F (.08 m) = 0$

This is the first equation relating F and v, now we need to use the second case to find another. In the second case momentum is conserved during the collision, so

1ex (.007 kg) v  =  (1.007 kg) v_f

so, solving for v_f, we get

$v_{f} ~=~ \frac{.007}{1.007} v$

again, looking at the change in kinetic energy in the second case yields

$K_{i} + \Delta K_{nc} ~=~ K_{f}$

$\frac{1}{2} (.007 kg) v^{2} - F d = \frac{1}{2} (1.007 kg) v_{f}^{2}$

substituting in our value for v_f that we found from the conservation of momentum,

$\frac{1}{2} (.007 kg) v^{2} - F d = \frac{1}{2} (1.007 kg) \left(\frac{.007}{1.007}\right)^{2} v^{2}$

we know have two equations involving d and v, so we can combine them. Simplification yields

$F d ~=~ F (.08 m) \left(1 - \frac{.007}{1.007}\right)$

and then dividing out the F on both sides yields d = 7.94 cm.