| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Collision then wall impact |
| Difficulty | Standard +0.8 This is a multi-stage collision problem requiring systematic application of conservation of momentum and restitution equations across two separate collisions. While the individual techniques are standard Further Maths mechanics, the problem requires careful tracking of directions, multiple unknowns, and extended calculation across 5 parts totaling 13 marks. The conceptual demand is moderate but the computational complexity and potential for sign errors elevates it above average difficulty. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03d Conservation in 2D: vector momentum6.03f Impulse-momentum: relation6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| 1(a) | \(e = 0.75\) | B1 |
| 1(b) | Conservation of momentum: \(18 \times 4 + 7 \times (-3) = 18v_A + 7v_B\) | M1 |
| All correct | A1 | |
| Restitution: \(v_B - v_A = -\frac{5}{7}(-3-4)\) | M1 | allow one sign error |
| All correct, any form | A1 | |
| \(18v_A + 7v_B = 51\) | ||
| \(-7v_A + 7v_B = 35\) | ||
| \(25v_A = 16\) | m1 | one variable eliminated |
| \(v_A = 0.64\) | A1 | cao |
| \(v_B = 5.64\) | A1 | cao |
| 1(c) | \(I = 7[5.64-(-3)]\) | M1 |
| \(I = 60.48\) Ns | A1 | ft \(v_A, v_B\) |
| 1(d) | Energy loss \(= 0.5(18 \times 4^2 + 7 \times 3^2) - 0.5(18 \times 0.64^2 + 7 \times 5.64^2)\) | M1 |
| \(= 60.48\) (J) | A1 | ft \(v_A, v_B\) provided answer +ve |
| 1(e) | After collision \(A\) moves towards the wall. | B1 |
1(a) | $e = 0.75$ | B1 | |
1(b) | **Conservation of momentum:** $18 \times 4 + 7 \times (-3) = 18v_A + 7v_B$ | M1 | allow 1 sign error |
| | All correct | A1 | |
| **Restitution:** $v_B - v_A = -\frac{5}{7}(-3-4)$ | M1 | allow one sign error |
| | All correct, any form | A1 | |
| $18v_A + 7v_B = 51$ | | |
| $-7v_A + 7v_B = 35$ | | |
| $25v_A = 16$ | m1 | one variable eliminated |
| $v_A = 0.64$ | A1 | cao |
| $v_B = 5.64$ | A1 | cao |
1(c) | $I = 7[5.64-(-3)]$ | M1 | oe, ft $v_A, v_B$ |
| $I = 60.48$ Ns | A1 | ft $v_A, v_B$ |
1(d) | Energy loss $= 0.5(18 \times 4^2 + 7 \times 3^2) - 0.5(18 \times 0.64^2 + 7 \times 5.64^2)$ | M1 | ft $v_A, v_B$ |
| $= 60.48$ (J) | A1 | ft $v_A, v_B$ provided answer +ve |
1(e) | After collision $A$ moves towards the wall. | B1 | ft $v_A$ |Two objects, $A$ of mass 18 kg and $B$ of mass 7 kg, are moving in the same straight line on a smooth horizontal surface. Initially, they are moving with the same speed of $4\text{ ms}^{-1}$ and in the same direction. Object $B$ collides with a vertical wall which is perpendicular to its direction of motion and rebounds with a speed of $3\text{ ms}^{-1}$. Subsequently, the two objects $A$ and $B$ collide directly. The coefficient of restitution between the two objects is $\frac{5}{7}$.
\begin{enumerate}[label=(\alph*)]
\item Find the coefficient of restitution between $B$ and the wall. [1]
\item Determine the speed of $A$ and the speed of $B$ immediately after the two objects collide. [7]
\item Calculate the impulse exerted by $A$ on $B$ due to the collision and clearly state its units. [2]
\item Find the loss in energy due to the collision between $A$ and $B$. [2]
\item State the direction of motion of $A$ relative to the wall after the collision with $B$. [1]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2018 Q1 [13]}}