Two forces F₁ and F₂ act on a particle P. The force F₁ is given by F₁ = (-i + 2j) N and F₂ acts in the direction of the vector (i + j). Given that the resultant of F₁ and F₂ acts in the direction of the vector (i + 3j), (a) find F₂ The acceleration of P is (3i + 9j) m s⁻². At time t = 0, the velocity of P is (3i - 22j) m s⁻¹. (b) Find the speed of P when t = 3 seconds.

A-Level Edexcel Maths Mechanics Kinematics Graphs: Two forces F₁ and F₂ act on a pa

Two forces F₁ and F₂ act on a particle P - Edexcel - A-Level Maths Mechanics - Question 7 - 2016 - Paper 1

Question 7

Two forces F₁ and F₂ act on a particle P. The force F₁ is given by F₁ = (-i + 2j) N and F₂ acts in the direction of the vector (i + j). Given that the resultant of... show full transcript

Worked Solution & Example Answer:Two forces F₁ and F₂ act on a particle P - Edexcel - A-Level Maths Mechanics - Question 7 - 2016 - Paper 1

Step 1

find F₂

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Answer

To find the force F₂, we first express the known forces:

The force F₁ is given as:

$F₁ = -i + 2j$
The direction of F₂ is the same as that of the vector (i + j). Hence, we can express F₂ as:

$F₂ = k(i + j)$
where k is a scalar.
The resultant force, given as the sum of F₁ and F₂, should act in the direction of (i + 3j):

$F_{resultant} = F₁ + F₂ = (-i + 2j) + k(i + j)$

Simplifying this gives:

$F_{resultant} = (-1 + k)i + (2 + k)j$
For the resultant to be in the direction of (i + 3j), the ratio of its components must equal:

$rac{-1 + k}{2 + k} = rac{1}{3}$
Cross-multiplying and simplifying, we have:

$3(-1 + k) = 1(2 + k)$

$-3 + 3k = 2 + k$

$2k = 5$

$k = 2.5$
Substituting k back into the expression for F₂:

$F₂ = 2.5(i + j) = 2.5i + 2.5j$ .

Step 2

Find the speed of P when t = 3 seconds

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Answer

To find the speed of P at t = 3 seconds, we first calculate the velocity after the specified time. The initial velocity is given as:

$v_0 = (3i - 22j) ext{ m s}^{-1}$

The acceleration of P is given by:

$a = (3i + 9j) ext{ m s}^{-2}$
We can use the equation of motion:

$v = v_0 + a t$
Substituting the values for t = 3 seconds:

$v = (3i - 22j) + (3i + 9j) imes 3$

Calculating the acceleration component:

$= (3i - 22j) + (9i + 27j)$

$= (3 + 9)i + (-22 + 27)j$

$= 12i + 5j ext{ m s}^{-1}$
Now, to find the speed, we determine the magnitude of the velocity vector:

$|v| = ext{sqrt}(12^2 + 5^2)$

$= ext{sqrt}(144 + 25) = ext{sqrt}(169) = 13 ext{ m s}^{-1}$

Thus, the speed of P when t = 3 seconds is 13 m s⁻¹.

Two forces F₁ and F₂ act on a particle P - Edexcel - A-Level Maths Mechanics - Question 7 - 2016 - Paper 1

Worked Solution & Example Answer:Two forces F₁ and F₂ act on a particle P - Edexcel - A-Level Maths Mechanics - Question 7 - 2016 - Paper 1

find F₂

Find the speed of P when t = 3 seconds

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