An archer shoots an arrow toward a 300 g target that is sliding in her direction at a speed of 2.15 m/s on a smooth, slippery surface. The 22.5 g arrow is shot with a speed of 38.0 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target?

Answer 1

THis can be found using the conservation of momentum.
Pix + Pix = Pfx + Pfx

Where P = mv

The mass of the arrow is .3 kg, the target is .0225kg. The initial velocity of the target is -2.15 m/s and the initial velocity of the arrow is 38.0m/s. If the target is stopped, the final velocity is 0 m/s. SO:

(.3kg)(38.0m/s) + (.0225kg)(-2.15m/s) = (.3kg)(Vf) + (.0225kg)(0m/s)

[(11.4Ns) + (-.048375Ns)]/(.3kg)=Vf

I'll let you figure it out from there. ;)

Answer 2

This is wrong. The 0m/s should be with the mass of the target, not of the arrow.