3(a)
Calculate the mass of the object using the formula:
$$ m = \frac{F}{a} $$
where the force (F) is 1870 N and the acceleration (a) is 1.83 m/s² - Edexcel - GCSE Physics Combined Science - Question 3 - 2016 - Paper 1
Question 3
3(a)
Calculate the mass of the object using the formula:
$$ m = \frac{F}{a} $$
where the force (F) is 1870 N and the acceleration (a) is 1.83 m/s².
3(b)
Using ... show full transcript
Worked Solution & Example Answer:3(a)
Calculate the mass of the object using the formula:
$$ m = \frac{F}{a} $$
where the force (F) is 1870 N and the acceleration (a) is 1.83 m/s² - Edexcel - GCSE Physics Combined Science - Question 3 - 2016 - Paper 1
Step 1
Calculate the mass
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Answer
To find the mass of the object, we will rearrange the formula for mass:
m=aF.
Substituting the given values:
Force (F) = 1870 N
Acceleration (a) = 1.83 m/s²
Thus,
m=1.831870≈1020 kg.
Rounding to three significant figures gives us 1020 kg.
Step 2
Calculate the final velocity
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Answer
For this part, rearrange the formula for final velocity:
v=u+at.
Here, the initial velocity (u) is assumed to be 0 m/s. Therefore:
a = 1.83 m/s²,
t = 16 s.
Substituting these values gives us:
v=0+1.83×16=29.28 m/s.
Rounding gives the final velocity as approximately 29.3 m/s.
Step 3
Calculate areas and distances
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Answer
For this part, you need to analyze the graph to identify the relevant data points and calculate areas.
The area under section AB is calculated to be 240 m².
The area under section CD calculates to 135 m².
Next, it should be noted:
The distance traveled at constant speed (240 m) is greater than the distance traveled when slowing down (135 m). Thus, this confirms that the object maintained a greater distance during its uniform motion.
