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 change of momentum of the arrow will be the same as the change of momentum of the target

initial target momentum = 0.300 * 2.35 = .705 kg m/s
final target momentum = 0
so the arrow momentum will be reduced by .705 kg m/s
final arrow momentum = .0225*41 - .705
= .2175 kg m/s
final arrow speed = .2175/.0225 = 9.67 m/s

To find the speed of the arrow after passing through the target, we can apply the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.

The momentum of an object is given by the product of its mass and its velocity. So, we can calculate the momentum of the arrow before the collision using its mass and initial velocity, and the momentum of the target using its mass and initial velocity.

Let's denote the mass of the arrow as m1 (22.5 g = 0.0225 kg) and the mass of the target as m2 (300 g = 0.3 kg).

The momentum of the arrow before the collision can be calculated as:
p1 = m1 * v1

where v1 is the initial velocity of the arrow (41.0 m/s).

Similarly, the momentum of the target before the collision is:
p2 = m2 * v2

where v2 is the initial velocity of the target (2.35 m/s).

Since the collision causes the target to stop, its velocity after the collision (v2') will be zero. Therefore, the momentum of the target after the collision will be zero as well.

To find the velocity of the arrow after the collision (v1'), we can substitute the conservation of momentum equation:

p1 + p2 = p1' + p2'

Since p2' is zero, the equation becomes:

p1 + p2 = p1'

Simplifying and solving for v1':

m1 * v1 + m2 * v2 = m1 * v1'

Now we can plug in the known values:
m1 = 0.0225 kg
v1 = 41.0 m/s
m2 = 0.3 kg
v2 = 2.35 m/s

Substituting these values into the equation:

(0.0225 kg * 41.0 m/s) + (0.3 kg * 2.35 m/s) = (0.0225 kg * v1')

Solving the equation:

0.9225 kg*m/s + 0.705 kg*m/s = 0.0225 kg * v1'

1.6275 kg*m/s = 0.0225 kg * v1'

Dividing by 0.0225 kg:

72.2 m/s = v1'

Therefore, the speed of the arrow after passing through the target is 72.2 m/s.