| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Work done against resistance - penetration into material |
| Difficulty | Moderate -0.3 This is a straightforward application of work-energy principles with clearly stated values and a two-part structure that guides students through the solution. Part (a) requires simple multiplication (work = force × distance), while part (b) applies the work-energy theorem with given initial conditions to verify a stated answer. The question involves standard M2 content with no conceptual surprises, making it slightly easier than average but still requiring proper understanding of the work-energy relationship. |
| Spec | 6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| (a) work done = force × dist. = \(8000 \times 0.04 = 320\) J | M1 A1 | |
| (b) work done = change in KE = \(\frac{1}{2}m(v^2 - u^2)\) | M1 | |
| \(= \frac{1}{2} \times 0.025(v^2 - 200^2) \therefore v^2 - 40000 = -25600\) | M2 A1 | |
| \(v^2 = 14400 \therefore v = 120\) ms\(^{-1}\) | A1 | (7) |
**(a)** work done = force × dist. = $8000 \times 0.04 = 320$ J | M1 A1 |
**(b)** work done = change in KE = $\frac{1}{2}m(v^2 - u^2)$ | M1 |
$= \frac{1}{2} \times 0.025(v^2 - 200^2) \therefore v^2 - 40000 = -25600$ | M2 A1 |
$v^2 = 14400 \therefore v = 120$ ms$^{-1}$ | A1 | (7)\begin{enumerate}
\item A bullet of mass 25 g is fired directly at a fixed wooden block of thickness 4 cm and passes through it. When the bullet hits the block, it is travelling horizontally at $200 \mathrm {~ms} ^ { - 1 }$. The block exerts a constant resistive force of 8000 N on the bullet.\\
(a) Find the work done by the block on the bullet.
\end{enumerate}
By using the Work-Energy principle,\\
(b) show that the bullet emerges from the block with speed $120 \mathrm {~ms} ^ { - 1 }$.\\
\hfill \mbox{\textit{Edexcel M2 Q1 [7]}}