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Question 13
In a school experiment, a particle, of mass m kilograms, is released from rest at a point h metres above the ground. At the instant it reaches the ground, the part... show full transcript
Step 1
Answer
To derive the equation for the velocity of the particle just before it reaches the ground, we can use the equations of motion. Considering that the initial velocity (u) is 0 and the acceleration (a) is due to gravity (g), we have:
From the second equation of motion:
v^2 = u^2 + 2as
Substituting the known values:
v^2 = 0 + 2gh
Thus, we arrive at the equation:
v = √(2gh).
This confirms the relationship between the velocity of the particle just before impact and the height from which it is released.
Step 2
Answer
We have the values of h = 18 m and the measured velocity v = 20 m/s. We can use the derived equation to determine what the theoretical velocity should be:
Using g ≈ 9.8 m/s², we substitute into the equation:
Calculating this, we find:
Now, comparing the two values:
Since the measured velocity (20 m/s) is greater than the expected theoretical value (18.8 m/s), it indicates a discrepancy. Therefore, the teacher's claim that the machine measuring velocity is faulty is indeed correct as the measured value does not align with the expected value.
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