A particle P of mass 2 kg is moving under the action of a constant force F newtons. When t = 0, P has velocity (3i + 2j) m s⁻¹ and at time t = 4 s, P has velocity (15i - 4j) m s⁻¹. Find (a) the acceleration of P in terms of i and j, (b) the magnitude of F, (c) the velocity of P at time t = 6 s.

Photo AI

A particle P of mass 2 kg is moving under the action of a constant force F newtons - Edexcel - A-Level Maths Mechanics - Question 3 - 2007 - Paper 1

Question 3

A particle P of mass 2 kg is moving under the action of a constant force F newtons. When t = 0, P has velocity (3i + 2j) m s⁻¹ and at time t = 4 s, P has velocity (1... show full transcript

Worked Solution & Example Answer:A particle P of mass 2 kg is moving under the action of a constant force F newtons - Edexcel - A-Level Maths Mechanics - Question 3 - 2007 - Paper 1

Step 1

the acceleration of P in terms of i and j

96%

114 rated

Answer

To find the acceleration, we first calculate the change in velocity over time. The initial velocity at t = 0 is:

$v_0 = 3i + 2j$

The final velocity at t = 4 s is:

$v_f = 15i - 4j$

The change in velocity ( $riangle v$ ) is:

$riangle v = v_f - v_0 = (15i - 4j) - (3i + 2j) = (15 - 3)i + (-4 - 2)j = 12i - 6j$

The acceleration (a) can be calculated using the formula:

$a = \frac{\triangle v}{\triangle t}$

Substituting in the change in velocity and the time interval (4 s):

$a = \frac{12i - 6j}{4} = 3i - 1.5j$

Step 2

the magnitude of F

99%

104 rated

Answer

Using Newton's second law, we apply:

$F = ma$

Where m = 2 kg and we have just found:

a = 3i - 1.5j.

Thus, substituting:

$F = 2(3i - 1.5j) = 6i - 3j$

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

$|F| = \sqrt{(6)^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45}$

Therefore, the magnitude of F is:

$|F| \approx 6.71 N$

Step 3

the velocity of P at time t = 6 s

96%

101 rated

Answer

We know the acceleration and can find the velocity at t = 6 s using:

$v_f = v_0 + at$

Here, at t = 6 s, we have:

Initial velocity: $v_0 = 3i + 2j$

Acceleration: $a = 3i - 1.5j$

So, substituting:

$v_f = (3i + 2j) + (3i - 1.5j)(6)$

Calculating:

$v_f = 3i + 2j + (18i - 9j)$

Combining like terms yields:

$v_f = (3 + 18)i + (2 - 9)j = 21i - 7j$

Therefore, the velocity of P at t = 6 s is:

$21i - 7j \, (m/s)$

A particle P of mass 2 kg is moving under the action of a constant force F newtons - Edexcel - A-Level Maths Mechanics - Question 3 - 2007 - Paper 1

Worked Solution & Example Answer:A particle P of mass 2 kg is moving under the action of a constant force F newtons - Edexcel - A-Level Maths Mechanics - Question 3 - 2007 - Paper 1

the acceleration of P in terms of i and j

the magnitude of F

the velocity of P at time t = 6 s

Join the A-Level students using SimpleStudy...