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?
(a) Because the arrow's deceleration and block's acceleration are acting in the same direction, add them to get a total acceleration of the arrow+styrofoam to be (-1300m/s2)+(-450m/s2)=-1750m/s2. Using v=v0+at where v=0m/s (final velocity of arrow relative to block), v0= 20m/s, you get t=0.0114s
(b) The styrofoam block starts from v0=0m/s, has an acceleration of a=450m/s2, and t from part (a)=0.0114s, we plug into v=v0+at to find v=5.13m/s
(c) Using x=v0t+.5at2, v0=0m/s because the initial velocity of the arrow compared to the block is insignificant, a=1750m/s2 from part (a), and t=0.0114s from part (a) to give us x=0.1137m
To solve this problem, we can use the equation of motion:
v² = u² + 2as
Where:
v = final velocity (0 m/s when the arrow stops moving)
u = initial velocity (20.0 m/s)
a = acceleration (1300 m/s² for the arrow, -450 m/s² for the block)
s = distance traveled into the block
(a) To find the time it takes for the arrow to stop moving, we can rearrange the equation:
0² = (20.0 m/s)² + 2(-1300 m/s²)s
0 = 400 m²/s² - 2600 m/s²s
2600 m/s²s = 400 m²/s²
s = (400 m²/s²) / (2600 m/s²)
s = 0.1538 m
Now, we can find the time using the formula:
v = u + at
0 = 20.0 m/s + (-1300 m/s²)t
-1300 m/s²t = -20.0 m/s
t = (-20.0 m/s) / (-1300 m/s²)
t = 0.0154 s
So, it takes approximately 0.0154 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 can consider the block's acceleration:
v = u + at
v = 0 m/s + (-450 m/s²)(0.0154 s)
v = -6.93 m/s
So, the common speed when the arrow stops is approximately 6.93 m/s.
(c) The distance into the block the arrow penetrates is given by:
s = ut + 0.5at²
s = (20.0 m/s)(0.0154 s) + 0.5(-1300 m/s²)(0.0154 s)²
s = 0.307 m
Therefore, the arrow penetrates approximately 0.307 meters into the block.
To find the solutions to the given problem, we can use the laws of physics, specifically the equations of motion and Newton's second law of motion. Let's break down each part of the question one by one.
(a) How long does it take for the arrow to stop moving with respect to the block?
We know that the arrow's deceleration has a magnitude of 1300 m/s^2. The deceleration is given by the equation:
deceleration = (final velocity - initial velocity) / time
We can rearrange this equation to solve for time:
time = (final velocity - initial velocity) / deceleration
Given that the initial velocity of the arrow is 20.0 m/s and the deceleration is 1300 m/s^2, we can substitute these values into the equation to find the time:
time = (0 - 20.0) / (-1300)
time = 0.0154 s
Therefore, 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?
To find the common speed of the arrow and block when they come to rest, we can use the concept of relative motion. When the arrow stops moving with respect to the block, their velocities are equal. The velocity of the block is given by the equation:
velocity_block = velocity_arrow + velocity_relative
where velocity_relative is the relative velocity between the arrow and the block.
Since the arrow comes to rest relative to the block, velocity_arrow is zero.
So, velocity_block = 0 + velocity_relative
Given that the magnitude of the block's acceleration is 450 m/s^2 and the time taken for the arrow to stop moving is 0.0154 s, we can find the relative velocity between the arrow and the block using the equation:
velocity_relative = acceleration_block * time
velocity_relative = 450 * 0.0154
velocity_relative = 6.93 m/s
Therefore, the common speed of the arrow and block when they come to rest is approximately 6.93 m/s.
(c) How far into the block does the arrow penetrate?
To find the distance the arrow penetrates into the block, we can use the equation of motion:
distance = (initial velocity * time) + (1/2 * acceleration * time^2)
Given that the initial velocity of the arrow is 20.0 m/s, the time taken for the arrow to stop moving is 0.0154 s, and the magnitude of the arrow's deceleration is 1300 m/s^2, we can substitute these values into the equation:
distance = (20.0 * 0.0154) + (0.5 * (-1300) * 0.0154^2)
distance = 0.308 + -1.51036
distance = -1.20236 m
The negative sign indicates that the arrow penetrated into the block in the opposite direction of its initial motion. Taking the absolute value, the arrow penetrates approximately 1.20 meters into the block.
Therefore, the arrow penetrates approximately 1.20 meters into the block.