An archer shoots an arrow toward a 300 g target that is sliding in her direction at a speed of 2.35 m/s on a smooth, slippery surface. The 22.5 g arrow is shot with a speed of 41.0 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?

The target loses momentum in the opposite direction to the arrow's motion, by an amount 300 g*2.35 m/s = 705 g m/s.

The arrow loses momentum equal to that, in the direction of its motion. Divide 705 g m/s by 22.5 g to get the decrease in the arrow's velocity. Then subtract that from the inital velocity.

Or just solve this equation:
Arrow mass*Initial arrow speed
-Target mass * Initial target speed = Arrow mass*Final arrow speed

To find the speed of the arrow after passing through the target, we can make use of the conservation of linear momentum principle. According to this principle, the total momentum of a system before and after a collision remains constant if no external forces act on it.

The momentum of an object is given by the product of its mass and velocity. The initial momentum of the system (including both the arrow and the target) is the sum of the individual momenta.

Before the collision:
Momentum of the arrow = mass of the arrow * velocity of the arrow
= 22.5 g * 41.0 m/s

Momentum of the target = mass of the target * velocity of the target
= 300 g * (-2.35 m/s) [The negative sign denotes the opposite direction]

Total initial momentum = Momentum of the arrow + Momentum of the target

Now, since the target is stopped by the impact, its final velocity is 0 m/s. Therefore, the final momentum of the target is 0.

After the collision:
Total final momentum = Momentum of the arrow + Momentum of the target (which is 0)

Since momentum is conserved, the initial and final total momenta are equal:

Total initial momentum = Total final momentum

22.5 g * 41.0 m/s + 300 g * (-2.35 m/s) = 0

Now, we can solve this equation to find the speed of the arrow after passing through the target.