Final answer:
The target's velocity after the arrow passes through it is 0.6 m/s, when a 0.1 kg arrow is shot from a bow at 60 m/s. The arrow hits a stationary 5 kg target and exits with a velocity of 30 m/s.
Step-by-step explanation:
When the arrow hits the stationary target, momentum is conserved.
We can apply the principle of conservation of momentum to solve for the target's velocity after the arrow passes through it.
First, we calculate the initial momentum of the arrow and the target:
Initial momentum of arrow = mass of arrow x initial velocity of arrow = 0.1 kg x 60 m/s
= 6 kg m/s
Initial momentum of target = mass of target x 0 (since it is stationary) = 5 kg x 0
= 0 kg m/s
Next, we calculate the final momentum of the arrow and the target:
Final momentum of arrow = mass of arrow x final velocity of arrow = 0.1 kg x 30 m/s
= 3 kg m/s
Final momentum of target = mass of target x final velocity of target
Using the principle of conservation of momentum, we equate the initial and final momentum:
Initial momentum of arrow + Initial momentum of target = Final momentum of arrow + Final momentum of target
6 kg m/s + 0 kg m/s = 3 kg m/s + Final momentum of target
Final momentum of target = 6 kg m/s - 3 kg m/s
= 3 kg m/s
Finally, we plug in the values to calculate the final velocity of the target:
Final momentum of target = mass of target x final velocity of target
3 kg m/s = 5 kg x final velocity of target
Final velocity of target = 3 kg m/s ÷ 5 kg
= 0.6 m/s
Therefore, the target's velocity after the arrow passes through it is 0.6 m/s.