A small stone is projected with speed 65 ms⁻¹ from a point O at the top of a vertical cliff. Point O is 70 m vertically above the point N. Point N is on horizontal ground. The stone is projected at an angle a above the horizontal, where tan a = \frac{5}{12}. The stone hits the ground at the point A, as shown in Figure 3. The stone is modelled as a particle moving freely under gravity. The acceleration due to gravity is modelled as having magnitude 10 ms⁻². Using the model, (a) find the time taken for the stone to travel from O to A, (b) find the speed of the stone at the instant just before it hits the ground at A. (c) State one limitation of the model that could affect the reliability of your answers.

Photo AI

A small stone is projected with speed 65 ms⁻¹ from a point O at the top of a vertical cliff - Edexcel - A-Level Maths Mechanics - Question 4 - 2021 - Paper 1

Question 4

A small stone is projected with speed 65 ms⁻¹ from a point O at the top of a vertical cliff. Point O is 70 m vertically above the point N. Point N is on horizontal... show full transcript

Worked Solution & Example Answer:A small stone is projected with speed 65 ms⁻¹ from a point O at the top of a vertical cliff - Edexcel - A-Level Maths Mechanics - Question 4 - 2021 - Paper 1

Step 1

find the time taken for the stone to travel from O to A

96%

114 rated

Answer

To find the time taken, we use the equation for vertical motion:

$s = ut + \frac{1}{2} a t^2$

Here, the initial vertical position is -70 m (since the stone falls 70 m), the initial vertical velocity component is (u = 65 \sin(a)), and the acceleration due to gravity is (a = -10 , \text{ms}^{-2}).

Substituting these values, we have:

$-70 = 65 \sin(a) \cdot t - \frac{1}{2} \cdot 10 t^2$

Given (\tan a = \frac{5}{12}), we can find (\sin a = \frac{5}{13}) and (\cos a = \frac{12}{13}).

Thus, substitute (\sin(a)):

$-70 = 65 \cdot \frac{5}{13} \cdot t - 5t^2$

By solving this quadratic equation, we can find the value of (t). The solution gives us approximately (t = 7 \text{ seconds}).

Step 2

find the speed of the stone at the instant just before it hits the ground at A.

99%

104 rated

Answer

The stone has both horizontal and vertical velocity components just before it hits the ground.

The horizontal component is given by: $v_x = 65 \cos(a)$ Substituting (\cos(a) = \frac{12}{13}):

$v_x = 65 \cdot \frac{12}{13} = 60 \, \text{ms}^{-1}$

The vertical component just before hitting the ground can be found using: $v_y = u_y + at$ Where (u_y = 65 \sin(a)) and substituting for (a = -10 , \text{ms}^{-2}):

$v_y = 65 \cdot \frac{5}{13} - 10 \cdot 7$

Calculating this gives: $v_y = 25 - 70 = -45 \, \text{ms}^{-1}$

The resultant speed is computed using Pythagoras’ theorem:

$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(60)^2 + (-45)^2} = \sqrt{3600 + 2025} = \sqrt{5625} \approx 75 \, \text{ms}^{-1}$

Step 3

State one limitation of the model that could affect the reliability of your answers.

96%

101 rated

Answer

One limitation of the model is that it ignores air resistance, which can affect the motion of the stone and lead to discrepancies between the model's predictions and real-world results. Other factors like wind effects and shape variations of the stone can also influence the accuracy of the model.

A small stone is projected with speed 65 ms⁻¹ from a point O at the top of a vertical cliff - Edexcel - A-Level Maths Mechanics - Question 4 - 2021 - Paper 1

Worked Solution & Example Answer:A small stone is projected with speed 65 ms⁻¹ from a point O at the top of a vertical cliff - Edexcel - A-Level Maths Mechanics - Question 4 - 2021 - Paper 1

find the time taken for the stone to travel from O to A

find the speed of the stone at the instant just before it hits the ground at A.

State one limitation of the model that could affect the reliability of your answers.

Join the A-Level students using SimpleStudy...