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1. A student carries out an experiment with a tennis ball and a motion sensor connected to a laptop - Scottish Highers Physics - Question 1 - 2019
Question 1
1. A student carries out an experiment with a tennis ball and a motion sensor connected to a laptop. The ball is released from rest below the sensor: The graph sho... show full transcript
Worked Solution & Example Answer:1. A student carries out an experiment with a tennis ball and a motion sensor connected to a laptop - Scottish Highers Physics - Question 1 - 2019
Step 1
Show that the initial acceleration of the ball is –9.8 m s⁻²
Answer
To find the initial acceleration, we use the formula: Here, from the graph, the change in velocity (\Delta v) between 0 s and 0.5 s is from 0 to -4.9 m/s. Hence, [\Delta v = -4.9 - 0 = -4.9 \text{ m/s}] The change in time (\Delta t = 0.5 - 0 = 0.5) s.
Now, substituting into the formula: [a = \frac{-4.9}{0.5} = -9.8 \text{ m/s}^2] Therefore, the initial acceleration of the ball is -9.8 m/s².
Step 2
Determine the height from which the ball is released.
Answer
To determine the height, we can use the equation of motion: [s = ut + \frac{1}{2}at^2] Where:
- (u = 0) (initial velocity, as it is released from rest)
- (a = -9.8 , \text{m/s}^2) (acceleration due to gravity)
- (t = 0.5) s (time interval)
Substituting the values: [s = 0 + \frac{1}{2}(-9.8)(0.5^2)] [s = \frac{1}{2}(-9.8)(0.25) = -1.225 , ext{m}]
Thus, the positive height from which the ball is released is 1.225 m.
Step 3
Determine the magnitude of the change in momentum of the ball during the bounce.
Answer
The change in momentum (\Delta p) can be calculated using: [\Delta p = mv - mu] Where:
- Mass (m = 0.057 , \text{kg}) (since 57.0 g = 0.057 kg)
- Initial velocity just before impact (u = -4.9 , \text{m/s}) (velocity at the lowest point)
- Final velocity after the bounce (use positive value) (v = 0.77 , \text{m/s})
Now calculating: [\Delta p = 0.057(0.77) - 0.057(-4.9)] [= 0.057(0.77 + 4.9) = 0.057 \times 5.67 = 0.32319 , ext{kg m/s}]
Thus, the magnitude of the change in momentum is approximately 0.3232 kg m/s.
Step 4
Determine the magnitude of the average force exerted by the ball on the ground during the bounce.
Answer
The average force (F) can be calculated using: [F = \frac{\Delta p}{\Delta t}] Where:
- (\Delta p \approx 0.3232 , \text{kg m/s}) (change in momentum)
- (\Delta t = 0.27 , \text{s}) (time taken during the bounce, not explicitly given but typically in the range of bounce times)
Substituting the values: [F = \frac{0.3232}{0.27} \approx 1.1963 , \text{N}]
Therefore, the magnitude of the average force exerted by the ball on the ground is approximately 1.20 N.
Step 5
Complete the sketch graph of acceleration a against time t for the ball, between 0 s and 1.18 s after it is released.
Answer
To complete the graph:
- Between 0 s and 0.5 s, the acceleration is constant at -9.8 m/s² (downward). This is represented by a horizontal line at -9.8 on the acceleration axis.
- From 0.5 s to 0.77 s, the ball experiences a uniform upward acceleration, transitioning back to 0.
- Following that, from 0.77 s to 1.18 s, the acceleration would drop back to 0 before rebounding.
A sketch of these intervals would show two lines:
- A line from (0, -9.8) to (0.5, -9.8).
- A line rising from (0.5, -9.8) to (0.77, 0) and then remaining at 0 until (1.18, 0).