First, we need to determine the distance the meteorite traveled before coming to a stop, which is half of the depth of the dent. This is because the meteorite would have been decelerated to the point of stopping and then accelerated in the opposite direction due to the collision:
distance = depth / 2 = 22 cm / 2 = 11 cm = 0.11 m
Next, we need to use the following kinematic equation:
v^2 - u^2 = 2as
where v = final velocity = 0 m/s (since the meteorite came to a stop), u = initial velocity = 550 m/s, s = distance = 0.11 m, and a = acceleration.
Rearranging the formula to find acceleration:
a = (v^2 - u^2) / 2s
Plugging in the values:
a = (0^2 - 550^2) / (2 × 0.11)
a = (-302500) / 0.22
a = -1375000 m/s^2
Since acceleration is a vector, we are only interested in the magnitude, not the direction, the acceleration is just:
|a| = 1375000 m/s^2.
On October 9, 1992 a 12.3 kg meteorite struck a car in Peekskill, NY; leaving a dent 22 cm deep in the trunk. If the meteorite struck the car with a speed of 550 m/s, what was the magnitude of the acceleration (assuming it was constant)?
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