Question

An arrow with a mass of 0.0180 kg enters a block of foam traveling at 138 ms and comes to a stop in 0.0042 s. The arrow entered into the block at a distance of 0.2898 m. What is the acceleration?

Answer 1

`a = 32857.167\,(m)/(s^(2))`

Step-by-step explanation:

The motion of the arrow can be described by the Moment Conservation and Impact Theorem:

`(0.018\,kg)\cdot \left(138\,(m)/(s) \right) - F\cdot (0.0042\,s) = (0.018\,kg)\cdot \left(0\,(m)/(s) \right)`

The impact force is computed herein:

`F = 591.429\,N`

The deceleration experimented by the arrow is:

`a = (F)/(m)`

`a = (591.429\,N)/(0.018\,kg)`

`a = 32857.167\,(m)/(s^(2))`