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A particle P of mass 0.4 kg moves under the action of a single constant force F newtons - Edexcel - A-Level Maths Mechanics - Question 3 - 2008 - Paper 1

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A particle P of mass 0.4 kg moves under the action of a single constant force F newtons. The acceleration of P is (6i + 8j) m s² . Find (a) the angle between the ac... show full transcript

Worked Solution & Example Answer:A particle P of mass 0.4 kg moves under the action of a single constant force F newtons - Edexcel - A-Level Maths Mechanics - Question 3 - 2008 - Paper 1

Step 1

the angle between the acceleration and i

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Answer

To find the angle θ between the acceleration vector and the i direction, we can use the tangent function. The acceleration vector is given as:

a=6i+8ja = 6i + 8j

Thus,

an(θ)=86 an(θ) = \frac{8}{6}
θ=tan1(86)θ = \tan^{-1}(\frac{8}{6})

Calculating this yields:

θ=53θ = 53^{\circ}

Step 2

the magnitude of F

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Answer

We can find the force F using Newton's second law, which states that:

F=maF = ma

Here, the mass m = 0.4 kg and the acceleration a = (6i + 8j) m s².

Calculating the force:

F=0.4(6i+8j)=2.4i+3.2jF = 0.4(6i + 8j) = 2.4i + 3.2j

To find the magnitude of F, we use the formula:

F=(2.4)2+(3.2)2|F| = \sqrt{(2.4)^2 + (3.2)^2}

Calculating this:

F=5.76+10.24=16=4|F| = \sqrt{5.76 + 10.24} = \sqrt{16} = 4

Step 3

find the velocity of P when t = 5

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Answer

The velocity of P at time t is given by:

v=9i10j+(6i+8j)tv = 9i - 10j + (6i + 8j)t

Substituting t = 5:

v=9i10j+(6i+8j)5v = 9i - 10j + (6i + 8j) \cdot 5

This simplifies to:

v=9i10j+(30i+40j)v = 9i - 10j + (30i + 40j)

Combining the terms:

v=(9+30)i+(10+40)j=39i+30jv = (9 + 30)i + (-10 + 40)j = 39i + 30j

Thus, the velocity of P when t = 5 is:

v=39i+30jext(ms1)v = 39i + 30j ext{ (m s}^{-1}\text{)}

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