2024 Edexcel IAS-Level Mathematics (XMA01) 模擬試題連答案詳解

(a) During the acceleration phase:
\( v = u + at_1 \implies v = 0 + 1.5T = 1.5T \)

During the deceleration phase:
\( v = u + at_3 \implies 0 = 1.5T - 2t_3 \implies t_3 = 0.75T \)

The total time taken for the journey is:
\( t_{\text{total}} = t_1 + t_2 + t_3 = T + 3T + 0.75T = 4.75T \) seconds.

(b) The total distance is the area under the speed-time graph:
\( S_1 = \frac{1}{2} \times T \times 1.5T = 0.75T^2 \)
\( S_2 = 3T \times 1.5T = 4.5T^2 \)
\( S_3 = \frac{1}{2} \times 0.75T \times 1.5T = 0.5625T^2 \)

Total distance:
\( S = S_1 + S_2 + S_3 = (0.75 + 4.5 + 0.5625)T^2 = 5.8125T^2 \)

Given \( S = 2325 \):
\( 5.8125T^2 = 2325 \implies T^2 = 400 \implies T = 20 \) (since \( T > 0 \)).

(c) The distance travelled during deceleration is \( S_3 \):
\( S_3 = 0.5625T^2 = 0.5625(400) = 225 \text{ m} \).

(a) M1: Attempts to find the deceleration time in terms of T using v = u + at.
A1: Correct deceleration time 0.75T.
A1: Adds the three times to show 4.75T (c.s.o.).

(b) M1: Sets up an expression for the total distance as the sum of three areas or using the trapezium formula.
A1: Correct simplified expression for the total distance, e.g., 5.8125 T^2.
M1: Equates their expression to 2325 and solves for T.
A1: T = 20.

(c) M1: Uses their value of T to calculate the distance during deceleration.
A1: 225 (m).

(a) At maximum height, the velocity of the stone is \( v = 0 \).
Using \( v = u + at \) with upwards as positive:
\( 0 = U - 9.8(2.5) \)
\( U = 24.5 \text{ m s}^{-1} \)

(b) Taking \( A \) as the origin, displacement \( s = -H \) when the stone hits the ground at \( t = 6 \):
Using \( s = ut + \frac{1}{2}at^2 \):
\( -H = 24.5(6) - \frac{1}{2}(9.8)(6^2) \)
\( -H = 147 - 176.4 = -29.4 \)
\( H = 29.4 \text{ m} \)

(c) Using \( v = u + at \) at \( t = 6 \):
\( v = 24.5 - 9.8(6) = 24.5 - 58.8 = -34.3 \text{ m s}^{-1} \)
The speed is the magnitude of the velocity:
\( \text{Speed} = 34.3 \text{ m s}^{-1} \)

(a) M1: Uses v = u + at with v = 0, t = 2.5 and g = 9.8.
A1: Correct equation 0 = U - 9.8(2.5).
A1: U = 24.5.

(b) M1: Applies s = ut + 0.5 a t^2 with t = 6, u = 24.5, a = -9.8.
A1: Correct substitution.
M1: Solves for H.
A1: H = 29.4 (or 29).

(c) M1: Uses v = u + at or v^2 = u^2 + 2as to find the velocity/speed at t = 6.
A1: Velocity of -34.3.
A1: Speed = 34.3 (or 34).

Given \( \tan\alpha = \frac{3}{4} \), we have \( \sin\alpha = 0.6 \) and \( \cos\alpha = 0.8 \).

(a) Resolving forces perpendicular to the plane:
\( R = mg\cos\alpha = 4(9.8)(0.8) = 31.36\text{ N} \)

Since \( P \) is on the point of sliding up the plane, the maximum friction force acts down the plane:
\( F_{\text{max}} = \mu R = 0.5(31.36) = 15.68\text{ N} \)

Resolving forces parallel to the plane (up the plane):
\( X - mg\sin\alpha - F_{\text{max}} = 0 \)
\( X = 4(9.8)(0.6) + 15.68 = 23.52 + 15.68 = 39.2\text{ N} \)

(b) When \( X \) is removed, the component of weight acting down the plane is:
\( W_{\parallel} = mg\sin\alpha = 23.52\text{ N} \)
Since \( W_{\parallel} = 23.52\text{ N} > F_{\text{max}} = 15.68\text{ N} \), the net force down the plane is positive, so the particle will slide down.

The equation of motion down the plane is:
\( mg\sin\alpha - F_{\text{max}} = ma \)
\( 23.52 - 15.68 = 4a \)
\( 7.84 = 4a \implies a = 1.96\text{ m s}^{-2} \)

(a) B1: Correct values of sin alpha = 0.6 and cos alpha = 0.8.
M1: Resolves perpendicular to the plane to find R.
A1: R = 31.36 (or 31.4) N.
M1: Resolves parallel to the plane with friction acting down the plane.
A1: X = 39.2 (or 39) N.

(b) M1: Compares the component of weight down the slope with maximum friction.
A1: Concludes correctly that P slides down since 23.52 > 15.68.
M1: Sets up the equation of motion down the plane.
A1: a = 1.96 (or 2.0) m s^-2.

(a) Let \( T_A \) and \( T_B \) be the tensions in \( AP \) and \( BP \) respectively.
Resolving forces horizontally:
\( T_A \cos 30^\circ = T_B \cos 45^\circ \)
\( T_A \frac{\sqrt{3}}{2} = T_B \frac{\sqrt{2}}{2} \)
\( T_B = T_A \frac{\sqrt{3}}{\sqrt{2}} = T_A \frac{\sqrt{6}}{2} \) (as required).

(b) Resolving forces vertically:
\( T_A \sin 30^\circ + T_B \sin 45^\circ = mg \)
Substitute \( m = 3 \), \( g = 9.8 \), and \( T_B = T_A \frac{\sqrt{6}}{2} \):
\( 0.5 T_A + \left( T_A \frac{\sqrt{6}}{2} \right) \frac{\sqrt{2}}{2} = 3(9.8) \)
\( 0.5 T_A + T_A \frac{\sqrt{12}}{4} = 29.4 \)
\( 0.5 T_A + T_A \frac{2\sqrt{3}}{4} = 29.4 \)
\( 0.5(1 + \sqrt{3}) T_A = 29.4 \)
\( T_A = \frac{58.8}{1 + \sqrt{3}} \approx 21.522 \approx 21.5\text{ N} \) (to 3 s.f.).

(c) Using the relation from part (a):
\( T_B = T_A \frac{\sqrt{6}}{2} \approx 21.522 \times 1.2247 \approx 26.359 \approx 26.4\text{ N} \) (to 3 s.f.).

(a) M1: Resolves horizontally with correct trig components.
A1: Obtains a correct horizontal equation.
A1: Shows clearly the step leading to the required exact relation.

(b) M1: Resolves vertically with both tension components and weight.
A1: Correct vertical equation.
M1: Eliminates T_B and solves for T_A.
A1: T_A = 21.5 (N) (accept 22).

(c) M1: Substitutes T_A into the relation from (a) or resolves to find T_B.
A1: T_B = 26.4 (N) (accept 26).

(a)(i) Position vector of \( A \) at time \( t \):
\( \mathbf{r}_A = (2 + 3t)\mathbf{i} + (5 - t)\mathbf{j} \)
Position vector of \( B \) at time \( t \):
\( \mathbf{r}_B = (-3 + ut)\mathbf{i} + (-7 + 2t)\mathbf{j} \)

At the point of collision, \( \mathbf{r}_A = \mathbf{r}_B \).
Comparing the \( \mathbf{j} \) components:
\( 5 - t = -7 + 2t \implies 3t = 12 \implies t = 4 \text{ s} \)

Comparing the \( \mathbf{i} \) components at \( t = 4 \):
\( 2 + 3(4) = -3 + 4u \)
\( 14 = -3 + 4u \implies 4u = 17 \implies u = 4.25 \)

(a)(ii) Substituting \( t = 4 \) into \( \mathbf{r}_A \):
\( \mathbf{r} = (2 + 12)\mathbf{i} + (5 - 4)\mathbf{j} = 14\mathbf{i} + \mathbf{j} \)

(b) If \( u = 5 \), then at \( t = 2 \):
\( \mathbf{r}_A = (2 + 3(2))\mathbf{i} + (5 - 2)\mathbf{j} = 8\mathbf{i} + 3\mathbf{j} \)
\( \mathbf{r}_B = (-3 + 5(2))\mathbf{i} + (-7 + 2(2))\mathbf{j} = 7\mathbf{i} - 3\mathbf{j} \)

The displacement vector from \( A \) to \( B \) is:
\( \vec{AB} = \mathbf{r}_B - \mathbf{r}_A = (7 - 8)\mathbf{i} + (-3 - 3)\mathbf{j} = -\mathbf{i} - 6\mathbf{j} \)

Distance:
\( |\vec{AB}| = \sqrt{(-1)^2 + (-6)^2} = \sqrt{37} \approx 6.08\text{ m} \) (to 3 s.f.).

(a)(i) M1: Writes general position vectors r_A and r_B in terms of t.
M1: Equates j-components and solves for collision time t.
A1: t = 4.
M1: Equates i-components at their t value and solves for u.
A1: u = 4.25.

(a)(ii) M1: Substitutes t = 4 into r_A or r_B.
A1: 14i + j.

(b) M1: Finds position vectors of A and B at t = 2 when u = 5.
M1: Finds the vector displacement between them.
A1: Distance = 6.08 (m) (or sqrt(37)).

(a) Let \( R_C \) and \( R_D \) be the reactions at \( C \) and \( D \) respectively.
Given \( R_C = 2R_D \).
Resolving vertically for equilibrium:
\( R_C + R_D = 12g \implies 3R_D = 12g \implies R_D = 4g \text{ N}, \, R_C = 8g \text{ N} \)

Taking moments about \( A \):
\( R_C(1) + R_D(4.5) = W(x) \)
\( 8g(1) + 4g(4.5) = 12g(x) \)
\( 8g + 18g = 12gx \implies 26g = 12gx \)
\( x = \frac{26}{12} = \frac{13}{6} \approx 2.17\text{ m} \) (to 3 s.f.).

(b) When the rod is on the point of tilting about \( D \), \( R_C = 0 \).
Taking moments about \( D \):
\( W(AD - x) = Mg(BD) \)
Where \( AD = 6 - 1.5 = 4.5\text{ m} \).
\( 12g \left(4.5 - \frac{13}{6}\right) = Mg(1.5) \)
\( 12 \left(\frac{27}{6} - \frac{13}{6}\right) = 1.5M \)
\( 12 \left(\frac{14}{6}\right) = 1.5M \)
\( 28 = 1.5M \implies M = \frac{28}{1.5} = \frac{56}{3} \approx 18.7\text{ kg} \) (to 3 s.f.).

(a) M1: Resolves vertically to find R_C and R_D in terms of g.
A1: Correct reactions R_D = 4g and R_C = 8g.
M1: Formulates a correct moments equation about any point (e.g. A, C or D).
A1: Correctly substituted moments equation.
M1: Solves for x.
A1: x = 13/6 or 2.17.

(b) M1: Identifies that reaction at C is zero when tilting about D.
M1: Formulates a correct moments equation about D with mass M at B.
A1: Correctly substituted moments equation.
A1: M = 18.7 (or 56/3).

(a) Let the initial direction of motion of \( P \) be the positive direction.
Initial velocities:
\( u_P = 3u \)
\( u_Q = -2u \)

Final velocities after collision:
\( v_P = -v \)
\( v_Q = 0.5u \)

Applying the principle of conservation of linear momentum:
\( m_P u_P + m_Q u_Q = m_P v_P + m_Q v_Q \)
\( 2m(3u) + 3m(-2u) = 2m(-v) + 3m(0.5u) \)
\( 6mu - 6mu = -2mv + 1.5mu \)
\( 0 = -2mv + 1.5mu \)
\( 2mv = 1.5mu \implies v = 0.75u \)

(b) Impulse \( I \) exerted on \( Q \):
\( I = m_Q(v_Q - u_Q) \)
\( I = 3m(0.5u - (-2u)) = 3m(2.5u) = 7.5mu \)

(c) The assumption that the surface is smooth means there is no friction between the surface and the particles. Consequently, there are no external horizontal forces acting on the system, which justifies the use of conservation of linear momentum.

(a) M1: Applies conservation of momentum with appropriate signs for velocities.
A1: Correct initial momentum side (6mu - 6mu = 0).
A1: Correct final momentum side (-2mv + 1.5mu).
M1: Equates and solves for v.
A1: v = 0.75u (or 3/4 u).

(b) M1: Uses Impulse = Change in momentum for Q.
A1: Correct calculation step, e.g., 3m(0.5u - (-2u)).
A1: 7.5mu.

(c) B1: Correctly explains that smooth surface means no friction/external horizontal forces, ensuring momentum is conserved.