| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Single particle, Newton's second law – vector (2D forces) |
| Difficulty | Moderate -0.3 This is a straightforward M1 mechanics question requiring basic vector addition and Newton's second law. Part (a) uses the condition that resultant force is parallel to motion direction (standard technique), and part (b) applies F=ma then v=u+at with given values. Both parts are routine applications of core mechanics principles with no problem-solving insight required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | (−3i+2j)+(pi+qj)= (−3+ p)i+(2+q)j | M1 |
| Answer | Marks |
|---|---|
| ( 2 + q ) − 2 | M1A1 |
| 2p+q−4=0* Allow 0=2p+q−4 but nothing else | A1* |
| Answer | Marks |
|---|---|
| 6(a) | M1 For adding and collecting i’s and j’s. |
| Answer | Marks | Guidance |
|---|---|---|
| 6(b) | p = 5 => q = − 6 => Resultant force = ( 2 i − 4 j ) | B1 |
| ( 2 i − 4 j ) = 0.5a | M1 | |
| v = (4i−8j)4 | M1 | |
| Speed = 1 6 2 + ( − 3 2 ) 2 = 1 2 8 0 = 1 6 5 = 36 ( m s − 1 ) or better | M1A1 |
| Answer | Marks |
|---|---|
| 6(b) | B1 Correct resultant force seen |
Question 6:
--- 6(a) ---
6(a) | (−3i+2j)+(pi+qj)= (−3+ p)i+(2+q)j | M1
( − 3 + p ) 1
=
( 2 + q ) − 2 | M1A1
2p+q−4=0* Allow 0=2p+q−4 but nothing else | A1*
(4)
6(a) | M1 For adding and collecting i’s and j’s.
N.B. Could be implied by p = 4 and q = -4
M1 Using ratios oe to set up an equation in p and q only, allow the
ratio the wrong way round.
M0 if they write down: −3+ p=1 and 2+q=−2 and NEVER use
ratios, but ignore these equations if they go on to use ratios
A1 Correct equation
A1* Correct answer correctly obtained
--- 6(b) ---
6(b) | p = 5 => q = − 6 => Resultant force = ( 2 i − 4 j ) | B1
( 2 i − 4 j ) = 0.5a | M1
v = (4i−8j)4 | M1
Speed = 1 6 2 + ( − 3 2 ) 2 = 1 2 8 0 = 1 6 5 = 36 ( m s − 1 ) or better | M1A1
(5)
(9)
Notes for question 6
6(b) | B1 Correct resultant force seen
M1 Use of F = ma OR F = ma where F (F) is their resultant (must
have attempted to add the two forces) (M0 if they include g)
M1 Use of v = at OR v = at with t = 4 where a or a is their
acceleration. (M0 if u o r u is non-zero)
M1 Use of Pythagoras to find magnitude of v OR a OR F, including
square root
N.B. The above 3 steps may appear in any order but must be entered on
ePen in the order as above.
A1 Any equivalent surd or correct to at least 2 SF[In this question, $\mathbf{i}$ and $\mathbf{j}$ are horizontal unit vectors.]
A particle $A$ of mass 0.5 kg is at rest on a smooth horizontal plane.
At time $t = 0$, two forces, $\mathbf{F}_1 = (-3\mathbf{i} + 2\mathbf{j})$ N and $\mathbf{F}_2 = (p\mathbf{i} + q\mathbf{j})$ N, where $p$ and $q$ are constants, are applied to $A$.
Given that $A$ moves in the direction of the vector $(\mathbf{i} - 2\mathbf{j})$,
\begin{enumerate}[label=(\alph*)]
\item show that $2p + q - 4 = 0$ [4]
Given that $p = 5$
\item Find the speed of $A$ at time $t = 4$ seconds. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2022 Q6 [9]}}