Question

An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.40 m/s on a smooth, slippery surface. the 22.5-g arrow is shot with a speed of 38.5 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

6.5 m/s
An object's momentum is its speed (m/s) multiplied by its mass(kg).
     initial momentum of target = 0.3kg x 2.4m/s = 0.72
   initial momentum of arrow = 0.0225kg x 38.5m/s = 0.86625
    resulting momentum of arrow is 0.86625 - 0.72 = 0.14625
    momentum /mass = speed
  so 0.14625 /0.0225 = 6.5 m/s

Answer 2

Answer:

The speed of the arrow after passing through the target is 6.5 m/s

Explanation:

Given that,

Mass of target = 300 g

Speed of target = 2.40 m/s

Mass of arrow = 22.5 g

Speed of arrow = 38.5 m/s

After impact the momentum of the target is zero since its velocity is zero.

After impact the velocity of the arrow can be determined with conservation of momentum

$\Delta p_{a}=-\Delta p_{t}$

We need to calculate the speed of the arrow after passing through the target

Using conservation of momentum

$m_{a}v_{a}_{i}+m_{t}v_{t}_{i}=-m_{a}v_{a}_{f}-m_{t}v_{t}_{f}$

Put the value into the formula

$0.0225\times38.5-0.3\times2.40=- 0.0225\times v_{f}-0$

$v_{f}=\dfrac{0.0225\times38.5-0.3\times2.40}{-0.0225}$

$v_{f}=6.5\ m/s$

Hence,The speed of the arrow after passing through the target is 6.5 m/s