***EXTRA CREDIT*** An arrow is fired with a speed of 20.0 m/s at a block of Styrofoam resting on a smooth surface. The arrow penetrates a certain distance into the block before coming to rest relative to it. During this process the arrow's deceleration has a magnitude of 1300 m/s2 and the block's acceleration has a magnitude of 450 m/s2
(a) How long does it take for the arrow to stop moving with respect to the block?
(b) What is the common speed of the arrow and block when this happens?
(c) How far into the block does the arrow penetrate?
To solve this problem, we need to use the equations of motion and the principles of kinematics. The equations of motion are:
1. v = u + at (equation of motion for linear motion)
2. v^2 = u^2 + 2as (equation of motion for linear motion)
Here:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time elapsed
- s is the displacement
Let's solve the problem step by step:
(a) How long does it take for the arrow to stop moving with respect to the block?
To find the time (t) for the arrow to stop moving with respect to the block, we can use the equation of motion (1) for the arrow's deceleration. Given that the arrow's initial velocity (u) is 20.0 m/s and the deceleration (a) is 1300 m/s^2, we can plug these values into the equation and solve for t:
0 = 20.0 - 1300t
Rearranging the equation, we get:
1300t = 20.0
t = 20.0 / 1300
t ≈ 0.0154 seconds
So, it takes approximately 0.0154 seconds for the arrow to stop moving with respect to the block.
(b) What is the common speed of the arrow and block when this happens?
When the arrow stops moving with respect to the block, their velocities become equal. To find this common speed, we need to find the final velocities of both the arrow and the block.
For the arrow: v_arrow = u_arrow + a_arrow * t
Given that u_arrow = 20.0 m/s, a_arrow = -1300 m/s^2 (negative sign indicates deceleration), and t = 0.0154 seconds (calculated in part (a)), we can calculate the final velocity of the arrow:
v_arrow = 20.0 + (-1300) * 0.0154
v_arrow ≈ -0.9692 m/s
For the block, we have:
v_block = 0 (since it comes to rest)
Therefore, the common speed of the arrow and block when they stop moving with respect to each other is approximately -0.9692 m/s.
(Note: The negative sign indicates that the arrow and block are moving in opposite directions at the time of impact.)
(c) How far into the block does the arrow penetrate?
To find the distance (s) that the arrow penetrates into the block, we can use the equation of motion (2) for the arrow's deceleration. Given that the initial velocity (u) is 20.0 m/s, the deceleration (a) is -1300 m/s^2, and the final velocity (v) is 0 m/s, we can plug these values into the equation and solve for s:
0^2 = 20.0^2 + 2 * (-1300) * s
0 = 400 - 2600s
2600s = 400
s = 400 / 2600
s ≈ 0.1538 meters
So, the arrow penetrates approximately 0.1538 meters into the block.