GCSE Edexcel Physics Combined Science Newtons Second Law: 3(a)
The formula for mass is giv

Question

3(a)
The formula for mass is given by:

$$m = \frac{F}{a}$$

where:
- $F$ is the force applied in Newtons,
- $a$ is the acceleration in m/s².

Using the values:  
$F = 1870$ N,  
$a = 1020$ kg.

Substituting these values into the formula:

$$m = \frac{1870}{1020}$$

This results in:

$$m \approx 1.83$$ kg,
rounded to three significant figures.

3(b)
To find the acceleration, we use the formula:

$\frac{v - u}{t} = a$

Where:
- $v$ is the final velocity,
- $u$ is the initial velocity (0 m/s),
- $t$ is time.

Given:
- $u = 0$ m/s,
- $a = 1.83$ m/s²,
- $t = 16$ s.

Substituting these values:

$$v = 0 + 1.83 	imes 16$$

Thus:

$$v \approx 29.3$$ m/s.

3(c)
From the graph, we can deduce several areas:

1. The area under the line segment AB is calculated to be:
   - Total area = 240 m.

2. For section CD:
   - Calculated area = 135 m.

3. When comparing distances:
   - The distance travelled at constant speed = 240 m,
   - This is greater than the distance travelled when slowing down = 135 m.

3(a) The formula for mass is given by: $$m = \frac{F}{a}$$ where: - $F$ is the force applied in Newtons, - $a$ is the acceleration in m/s² - Edexcel - GCSE Physics Combined Science - Question 3 - 2016 - Paper 1

Worked Solution & Example Answer:3(a) The formula for mass is given by: $$m = \frac{F}{a}$$ where: - $F$ is the force applied in Newtons, - $a$ is the acceleration in m/s² - Edexcel - GCSE Physics Combined Science - Question 3 - 2016 - Paper 1

3(a) Rearrangement of m = F/a

3(b) Substitution

3(c) Calculating areas and distances

Join the GCSE students using SimpleStudy...