Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.38 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.9 mm. If the floor is carpeted, this stopping distance is increased to about 1.2 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate.
To calculate the magnitude and duration of deceleration in both cases, we can use the equations of motion. Here's how you can approach the problem:
1. Calculate the initial velocity of the child's head just before it hits the floor in both cases.
a) For the case of hardwood floor:
The child falls a vertical distance of 0.38 m, so the initial velocity of the head can be found using the equation:
v^2 = u^2 + 2as
where v is the final velocity (0 m/s as the head comes to rest), u is the initial velocity, a is the acceleration (unknown), and s is the distance (0.38 m).
Rearrange the equation to solve for u:
u = sqrt(v^2 - 2as)
u = sqrt(0 - 2 * a * 0.38)
u = sqrt(-0.76a)
Note: Since the child falls vertically, there is no horizontal component of velocity to consider.
b) For the case of carpeted floor:
Follow the same process as before, but this time the distance fallen is 0.012 m (1.2 cm). Calculate the initial velocity using the equation:
u = sqrt(v^2 - 2as)
u = sqrt(0 - 2 * a * 0.012)
u = sqrt(-0.024a)
2. Calculate the deceleration (negative acceleration) in each case.
Using the stopping distance given, we can assume uniform deceleration during the head's deceleration.
a) For the case of hardwood floor:
The stopping distance is given as 1.9 mm (0.0019 m).
The deceleration can be calculated using the equation:
v^2 = u^2 + 2as
0^2 = u^2 + 2 * (-a) * 0.0019
0 = u^2 - 0.0038a
Substitute the value of u from step 1a:
0 = (-0.76a) - 0.0038a
0 = -0.7638a
Solving for a:
a = 0 m/s^2
Since the deceleration is 0 m/s^2, the head would experience no injury.
b) For the case of carpeted floor:
Follow the same procedure as before, but this time the stopping distance is 0.012 m.
0^2 = u^2 + 2 * (-a) * 0.012
0 = u^2 - 0.024a
Substitute the value of u from step 1b:
0 = (-0.024a) - 0.024a
0 = -0.048a
Solving for a:
a = 0 m/s^2
Similar to the previous case, the deceleration is 0 m/s^2, so there is no risk of injury.
In both cases, the deceleration is 0 m/s^2, which is significantly lower than the threshold of 1000 m/s^2 required to cause injury. Therefore, the child would not be at risk of injury in either case.