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?

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To solve this problem, we will use the equations of motion. Let's break down the problem step by step.

(a) To find the time it takes for the arrow to stop moving with respect to the block, we need to determine the deceleration time for the arrow. We can use the equation:

v = u + at

Where:
v = final velocity (0 m/s in this case, as the arrow comes to rest)
u = initial velocity (20.0 m/s in this case)
a = deceleration (-1300 m/s^2 in this case)
t = time

Rearranging the equation to solve for time:

t = (v - u) / a

Substituting the values:

t = (0 - 20.0) / (-1300)
t = 20.0 / 1300
t ≈ 0.015 s

Therefore, it takes approximately 0.015 seconds for the arrow to stop moving with respect to the block.

(b) To find the common speed of the arrow and block when the arrow stops, we need to determine the final velocity of the block. We can use the same equation as above:

v = u + at

Where:
v = final velocity of the block
u = initial velocity of the block (0 m/s in this case)
a = acceleration of the block (450 m/s^2 in this case)
t = time (0.015 s in this case)

Substituting the values:

v = 0 + (450) * (0.015)
v ≈ 6.75 m/s

Therefore, the common speed of the arrow and block when the arrow comes to rest is approximately 6.75 m/s.

(c) To find the distance the arrow penetrates into the block, we can use the equation of motion:

s = ut + (1/2)at^2

Where:
s = distance
u = initial velocity of the arrow (20.0 m/s in this case)
t = time (0.015 s in this case)
a = deceleration (-1300 m/s^2 in this case)

Substituting the values:

s = (20.0) * (0.015) + (1/2) * (-1300) * (0.015)^2
s ≈ 0.135 m

Therefore, the arrow penetrates approximately 0.135 meters into the block.

Note: It's important to keep track of the signs (positive and negative) when dealing with acceleration and deceleration.