Photo AI

A particle P of mass 4 kg is at rest at the point A on a smooth horizontal plane - Edexcel - A-Level Maths Mechanics - Question 3 - 2022 - Paper 1

Question icon

Question 3

A-particle-P-of-mass-4-kg-is-at-rest-at-the-point-A-on-a-smooth-horizontal-plane-Edexcel-A-Level Maths Mechanics-Question 3-2022-Paper 1.png

A particle P of mass 4 kg is at rest at the point A on a smooth horizontal plane. At time t = 0, two forces, F₁ = (4i - j) N and F₂ = (λi + μj) N, where λ and μ are... show full transcript

Worked Solution & Example Answer:A particle P of mass 4 kg is at rest at the point A on a smooth horizontal plane - Edexcel - A-Level Maths Mechanics - Question 3 - 2022 - Paper 1

Step 1

show that λ - 3μ + 7 = 0

96%

114 rated

Answer

To show that the forces result in the motion of P in the direction of the vector (3i + j), we first find the resultant force on P.

The forces in the i and j directions are:

  • For F₁: 4 N in the i direction and -1 N in the j direction.
  • For F₂: λ N in the i direction and μ N in the j direction.

So, the resultant force

Fextresultant=F1+F2=(4+λ)i+(1+μ)jF_{ ext{resultant}} = F₁ + F₂ = (4 + λ)i + (-1 + μ)j

Since P moves in the direction of (3i + j), we can express the relationship between the i and j components of the resultant force and the given motion vector using ratios:

4+λ3=1+μ1\frac{4 + λ}{3} = \frac{-1 + μ}{1}

From the first ratio,

4+λ=3λ=34=14 + λ = 3 \Rightarrow λ = 3 - 4 = -1

From the second ratio,

1+μ=1μ=1+1=2-1 + μ = 1 \Rightarrow μ = 1 + 1 = 2

Now substituting λ and μ values into the equation:

λ3μ+7=13(2)+7=16+7=0λ - 3μ + 7 = -1 - 3(2) + 7 = -1 - 6 + 7 = 0

Thus, we conclude:

λ3μ+7=0λ - 3μ + 7 = 0

Step 2

find the length of AB

99%

104 rated

Answer

Given λ = 2, we can now substitute this back into the expression for F₂. We also need to find the resultant force that dictates the motion of P:

F2=(2i+μj)F₂ = (2i + μj)

Now incorporating the given values:

Fextresultant=(4+2)i+(1+μ)j=(6)i+(1+μ)jF_{ ext{resultant}} = (4 + 2)i + (-1 + μ)j = (6)i + (-1 + μ)j

To find μ, first apply the previously found result:

Using the equation from part (a),

λ3μ+7=023μ+7=03μ=9μ=3λ - 3μ + 7 = 0 \Rightarrow 2 - 3μ + 7 = 0 \Rightarrow 3μ = 9 \Rightarrow μ = 3

The position at time t = 0 is still at point A, with a resultant force of:

Fextresultant=(6i+2j)F_{ ext{resultant}} = (6i + 2j)

Now, using the second equation of motion with initial conditions (u = 0) and t = 4,

s=ut+12at2s = ut + \frac{1}{2}at^2

Where we need to find the acceleration:

a=Fextresultantm=(6i+2j)4=(1.5i+0.5j)a = \frac{F_{ ext{resultant}}}{m} = \frac{(6i + 2j)}{4} = (1.5i + 0.5j)

Calculating the displacement:

s=0+12(1.5i+0.5j)(4)2=(12i+4j)s = 0 + \frac{1}{2}(1.5i + 0.5j)(4)^2 = (12i + 4j)

Therefore, point B from point A is just:

AB = ∥s∥ = (12)2+(4)2=144+16=160=410\sqrt{(12)^2 + (4)^2} = \sqrt{144 + 16} = \sqrt{160} = 4\sqrt{10}.

Thus the length of AB is:

410extm4\sqrt{10} ext{ m}

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;