A hockey puck moving at 37m/s slams through a wall of snow 31cm thick. It emerges moving at 15m/s .
Part A
Assuming constant acceleration, find the time the puck spends in the snow.
Express your answer to two significant figures and include the appropriate units.
Part B
Assuming constant acceleration, find the thickness of a snow wall that would stop the puck entirely.
Express your answer to two significant figures and include the appropriate units.
Part A:
To find the time the puck spends in the snow, we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity = 15 m/s
u = initial velocity = 37 m/s
a = acceleration (assumed to be constant)
s = displacement = 31 cm = 0.31 m
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Plugging in the given values:
0.31 = (15^2 - 37^2) / (2a)
Simplifying further:
0.31 = (225 - 1369) / (2a)
0.31 = -1144 / (2a)
To solve for 'a', we can rearrange the equation:
a = -1144 / (0.31 * 2) ≈ -1845.161 m/s^2
Since acceleration cannot be negative in this context, we take the magnitude of the acceleration:
|a| ≈ 1845.161 m/s^2
Now, using the equation of motion again:
v = u + at
Rearranging to solve for 't':
t = (v - u) / a
Plugging in the given values:
t = (15 - 37) / 1845.161
Calculating the result:
t ≈ -0.011 seconds
Since time cannot be negative in this context, we discard the negative sign. Therefore, the time the puck spends in the snow is approximately 0.011 seconds.
Part B:
To find the thickness of a snow wall that would stop the puck entirely, we can use the same equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity = 0 m/s (since the puck stops)
u = initial velocity = 37 m/s
a = acceleration (assumed to be constant)
s = unknown
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Plugging in the given values:
s = (0 - 37^2) / (2a)
Simplifying further:
s = (-1369) / (2a)
To solve for 's', we need to find the magnitude of the acceleration 'a'. From Part A, we found that |a| ≈ 1845.161 m/s^2.
Plugging in the value of 'a':
s = (-1369) / (2 * 1845.161)
Calculating the result:
s ≈ -0.37 m
Since the thickness of the snow wall cannot be negative, we discard the negative sign. Therefore, the thickness of a snow wall that would stop the puck entirely is approximately 0.37 meters.
To solve this problem, we can use the equations of motion for constant acceleration. We'll need to use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
For Part A:
Given:
Initial velocity (u) = 37 m/s
Final velocity (v) = 15 m/s
We need to find the time (t) the puck spends in the snow, so we'll rearrange the equation to solve for time:
v^2 = u^2 + 2as
15^2 = 37^2 + 2a(0.31)
Solving for acceleration (a):
2a(0.31) = 15^2 - 37^2
a = (15^2 - 37^2) / (2 * 0.31)
Next, we can use the equation v = u + at to find the time t:
15 = 37 + a * t
t = (15 - 37) / a
Substituting the value of a we found earlier:
t = (15 - 37) / [(15^2 - 37^2) / (2 * 0.31)]
Calculating this expression, we get:
t ≈ -2.60 seconds (rounded to two significant figures)
However, time cannot be negative, so the puck does not spend time in the snow. This means that the given values for the velocities might not be correct or that there is another factor we need to consider.
For Part B:
We need to find the thickness of a snow wall (s) that would stop the puck entirely.
Given:
Initial velocity (u) = 37 m/s
Final velocity (v) = 0 m/s (since the puck is stopped)
Acceleration (a) = ? (unknown)
Time (t) = ? (unknown)
Using the equation v^2 = u^2 + 2as and rearranging it to solve for displacement:
0^2 = 37^2 + 2a(s)
To stop the puck entirely, v = 0, so the equation becomes:
0 = 37^2 + 2a(s)
-37^2 = 2a(s)
Simplifying:
s = (-37^2) / (2a)
Now, we need to find the value of acceleration (a) to calculate the thickness of the snow wall. Given that the initial and final velocities are not provided, we cannot determine the acceleration, and therefore the thickness of the snow wall.
In conclusion:
For Part A, assuming constant acceleration, the puck does not spend time in the snow.
For Part B, assuming constant acceleration, we cannot determine the thickness of the snow wall without additional information.
V^2 = Vo^2 + 2a*d = 15^2
37^2 + 2a*0.31 = 15^2
2a*0.31 = 15^2 - 37^2
a = (15^2-37^2)/0.62 = -1845.2 m/s^2
A. V = Vo + a*t
t = (V-Vo)/a = (15-37)/-1845.2 = 0.012 s
B. V^2 = Vo^2 + 2a*d
V = 0
Vo = 37 m/s
a = -1845.2
Solve for d(thickness).