A stone S is sliding on ice. The stone is moving along a straight line ABC, where AB = 24 m and AC = 30 m. The stone is subject to a constant resistance to motion of magnitude 0.3 N. At A the speed of S is 20 m s⁻¹, and at B the speed of S is 16 m s⁻¹. Calculate (a) the deceleration of S, (b) the speed of S at C. (c) Show that the mass of S is 0.1 kg. At C, the stone S hits a vertical wall, rebounds from the wall and then slides back along the line CA. The magnitude of the impulse of the wall on S is 2.4 N s and the stone continues to move against a constant resistance of 0.3 N. (d) Calculate the time between the instant that S rebounds from the wall and the instant that S comes to rest.

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A stone S is sliding on ice - Edexcel - A-Level Maths Mechanics - Question 6 - 2005 - Paper 1

Question 6

A stone S is sliding on ice. The stone is moving along a straight line ABC, where AB = 24 m and AC = 30 m. The stone is subject to a constant resistance to motion of... show full transcript

Worked Solution & Example Answer:A stone S is sliding on ice - Edexcel - A-Level Maths Mechanics - Question 6 - 2005 - Paper 1

Step 1

Calculate the deceleration of S

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Answer

To find the deceleration of S while it is moving from A to B, we can use the following kinematic equation:

$v^2 = u^2 + 2as$

Where:

$v$ = final velocity at B = 16 m/s,
$u$ = initial velocity at A = 20 m/s,
$a$ = deceleration,
$s$ = distance between A and B = 24 m.

Substituting the values:

$16^2 = 20^2 + 2a(24)$

Solving for $a$ gives:

$256 = 400 + 48a$

Rearranging

$48a = 256 - 400$

$48a = -144$

Thus, the deceleration is:

$a = -3 ext{ m/s}^2$

Step 2

Calculate the speed of S at C

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Answer

To find the speed of S at point C, we will use the same kinematic equation:

$v^2 = u^2 + 2as$

This time, we use the initial speed at B = 16 m/s, deceleration = -3 m/s², and the distance from B to C = 30 m:

$v^2 = 16^2 + 2(-3)(30)$

Calculating:

$v^2 = 256 - 180$

$v^2 = 76$

Thus, the speed of S at C is:

$v = ext{sqrt}(76) ext{ or } 8.72 ext{ m/s}$

Step 3

Show that the mass of S is 0.1 kg

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Answer

Using the formula for the force of resistance:

$F = ma$

Where:

$F$ = 0.3 N (resistance),
$m$ = mass (unknown),
$a$ = acceleration = 3 m/s² from part (a).

Substituting the values:

$0.3 = m imes 3$

This leads to:

$m = \frac{0.3}{3} = 0.1 ext{ kg}$

Step 4

Calculate the time between the instant that S rebounds from the wall and the instant that S comes to rest

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Answer

At C, the stone rebounds with an impulse of 2.4 N s. The force acting on S after rebounding includes the resistance:

Calculate the net force:

$\text{Net Force} = \text{Impulse} - \text{Resistance}$

$\text{Net Force} = 2.4 - 0.3 = 2.1 ext{ N}$

Now, using Newton's second law:

$F = ma$

Where:

$F$ = 2.1 N,
$m$ = 0.1 kg,
$a = \frac{F}{m} = \frac{2.1}{0.1} = 21 ext{ m/s}^2$ (acceleration).

Considering S comes to rest, the deceleration will apply:

Using the formula:

$v = u + at$

Where:

$u = 0$ (final speed),
$a = -21$ m/s² (deceleration),
$v = 2.4$ m/s (speed just after rebounding).

Setting up the equation:

$0 = 2.4 - 21t$

Thus:

$21t = 2.4 \\ t = \frac{2.4}{21}$

Calculating:

$t \approx 0.1143 ext{ seconds}$ .

This is the time taken for S to come to a complete stop.

A stone S is sliding on ice - Edexcel - A-Level Maths Mechanics - Question 6 - 2005 - Paper 1

Worked Solution & Example Answer:A stone S is sliding on ice - Edexcel - A-Level Maths Mechanics - Question 6 - 2005 - Paper 1

Calculate the deceleration of S

Calculate the speed of S at C

Show that the mass of S is 0.1 kg

Calculate the time between the instant that S rebounds from the wall and the instant that S comes to rest

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