Photo AI

6 (a) The symbol 'g' can be used to refer to the acceleration due to gravity - Edexcel - GCSE Physics - Question 6 - 2018 - Paper 1

Question icon

Question 6

6-(a)-The-symbol-'g'-can-be-used-to-refer-to-the-acceleration-due-to-gravity-Edexcel-GCSE Physics-Question 6-2018-Paper 1.png

6 (a) The symbol 'g' can be used to refer to the acceleration due to gravity. The acceleration due to gravity 'g' has the unit of m/s². 'g' can also have another u... show full transcript

Worked Solution & Example Answer:6 (a) The symbol 'g' can be used to refer to the acceleration due to gravity - Edexcel - GCSE Physics - Question 6 - 2018 - Paper 1

Step 1

Which of these is also a unit for g?

96%

114 rated

Answer

The correct answer is C N/kg. To understand why, we consider that acceleration is defined as force per unit mass. Therefore, the unit of force is Newton (N), and dividing this by mass (kg) gives us N/kg, which is indeed a valid unit for acceleration.

Step 2

Calculate the average value of time t to an appropriate number of significant figures.

99%

104 rated

Answer

To calculate the average value of time t, we sum the three recorded values:

t = rac{0.74 + 0.69 + 0.81}{3}

Calculating this gives:

t = rac{2.24}{3} = 0.746666...

Rounding this to two significant figures gives an average value of time t = 0.75 s.

Step 3

Explain one way the students could improve their procedure to obtain a more accurate value for g.

96%

101 rated

Answer

One way students could improve their procedure is by conducting multiple trials and taking the average of a larger set of measurements. This helps to reduce random errors and provides a more reliable estimate of g.

Step 4

Calculate the deceleration of the car.

98%

120 rated

Answer

Using the equation of motion:

v2=u2+2asv^2 = u^2 + 2as

where:

  • v = final velocity = 0 m/s (the car comes to rest),
  • u = initial velocity = 15 m/s,
  • a = deceleration (which we want to find),
  • s = distance = 14 m.

Substituting the known values gives:

0=(15)2+2(a)(14)0 = (15)^2 + 2(a)(14)

Rearranging yields:

2a(14)=2252a(14) = -225

Thus:

ightarrow a = -8.04 ext{ m/s}^2$$ The deceleration of the car is approximately 8.04 m/s².

Join the GCSE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;