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A student carries out an experiment to investigate friction between a puck and the surface of a table - Scottish Highers Physics - Question 2 - 2022

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A student carries out an experiment to investigate friction between a puck and the surface of a table. The student pushes the puck and releases it at point R. The s... show full transcript

Worked Solution & Example Answer:A student carries out an experiment to investigate friction between a puck and the surface of a table - Scottish Highers Physics - Question 2 - 2022

Step 1

Calculate the average acceleration of the puck between point R and the centre of the target.

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Answer

To calculate the average acceleration (a), we can use the equation of motion:

a=v−uta = \frac{v - u}{t}

Where:

  • v = final velocity (0 m/s, since the puck comes to rest)
  • u = initial velocity (0.78 m/s)
  • d = distance travelled (2.160 m)

Using the equation:

v2=u2+2adv^2 = u^2 + 2ad

Rearranging for a yields:

a=v2−u22da = \frac{v^2 - u^2}{2d}

Substituting the known values:

a=0−(0.78)22×2.160a = \frac{0 - (0.78)^2}{2 \times 2.160}

a=−0.60844.32≈−0.140m/s2a = \frac{-0.6084}{4.32} \approx -0.140 m/s^2

Thus, the average acceleration of the puck is approximately -0.140 m/s².

Step 2

Calculate the magnitude of the average force of friction between the puck and the table.

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Answer

The average force of friction (F_f) can be calculated using Newton's second law:

Ff=m⋅aF_f = m \cdot a

Where:

  • m = mass of the puck (0.350 kg)
  • a = average acceleration (-0.140 m/s²)

Substituting the values:

Ff=0.350⋅(−0.140)F_f = 0.350 \cdot (-0.140)

Ff=−0.049 NF_f = -0.049 \text{ N}

The magnitude of the average force of friction is therefore approximately 0.049 N.

Step 3

Explain why the student’s statement is incorrect.

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Answer

The statement is incorrect because while improving the mass measurement precision will enhance the accuracy of the mass, it does not significantly affect the overall uncertainty in the calculation of the average force of friction. The average force of friction depends on both the mass and the acceleration due to friction. Reducing the uncertainty in mass alone does not address uncertainties in the velocity measurement or the distance travelled, so other aspects need to be improved simultaneously to significantly reduce the overall uncertainty in the friction force calculation.

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