| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Sphere rebounds off fixed wall obliquely |
| Difficulty | Standard +0.3 This is a straightforward application of momentum and impulse formulas in 2D. Part (a) requires calculating magnitude of momentum vector (basic vector arithmetic), (b) uses impulse = change in momentum, (c) applies Newton's third law, and (d) finds magnitude and angle from a vector. All steps are routine calculations with no problem-solving insight required, making it slightly easier than average but not trivial due to the multi-part vector work. |
| Spec | 6.02f KE with vectors: using scalar product6.03a Linear momentum: p = mv6.03e Impulse: by a force6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | p = mu = 12.5(–5i + 12j) |
| Answer | Marks |
|---|---|
| awrt 163 (kgms–1 ) | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Using definition of momentum soi |
| Answer | Marks |
|---|---|
| (b) | I = mv – mu = 12.5(i + 4j) – 12.5(–5i + 12j) |
| = 75i – 100j (Ns) | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Using I = mv – mu. |
| Answer | Marks | Guidance |
|---|---|---|
| ISW | Condone 12.5(6i -8j) oe | |
| (c) | –75i + 100j (Ns) | B1FT |
| [1] | 1.1 | FT – their answer to (b). |
| Answer | Marks |
|---|---|
| for units | Condone 12.5(-6i +8j) oe |
| (d) | I = (752 + 1002) = 125 (Ns) |
| Answer | Marks |
|---|---|
| rad | B1FT |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | FT – their answer to (b). |
| Answer | Marks |
|---|---|
| come from correct I | eg tanθ = 100/75 or 75/100 |
Question 1:
1 | (a) | p = mu = 12.5(–5i + 12j)
p = 12.5((–5)2 + 122) = 12.513 = 162.5
awrt 163 (kgms–1 ) | M1
A1
[2] | 1.1
1.1 | Using definition of momentum soi
e.g. implied by 12.5 x 13
Oe 325/2
(b) | I = mv – mu = 12.5(i + 4j) – 12.5(–5i + 12j)
= 75i – 100j (Ns) | M1
A1
[2] | 1.1
1.1 | Using I = mv – mu.
Could be in column vector form.
ISW | Condone 12.5(6i -8j) oe
(c) | –75i + 100j (Ns) | B1FT
[1] | 1.1 | FT – their answer to (b).
Vector form must be correct
(could be column form) No need
for units | Condone 12.5(-6i +8j) oe
(d) | I = (752 + 1002) = 125 (Ns)
I.i = 75 => cosθ = 75/125 = 3/5
θ = awrt -53.1° or -0.927 rad or 307° or 5.36
rad or +53.1° or +0.927 rad or -307° or -5.36
rad | B1FT
M1FT
A1
[3] | 1.1
1.1
1.1 | FT – their answer to (b).
Use of dot product, or trig, to
obtain a value for sin, cos or tan
Positive or negative values are
acceptable.
Cao. Mark final answer. Must
come from correct I | eg tanθ = 100/75 or 75/1001 A particle $P$ of mass 12.5 kg is moving on a smooth horizontal plane when it collides obliquely with a fixed vertical wall.
At the instant before the collision, the velocity of $P$ is $- 5 \mathbf { i } + 12 \mathbf { j } \mathrm {~ms} ^ { - 1 }$.\\
At the instant after the collision, the velocity of $P$ is $\mathbf { i } + 4 \mathbf { j } \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the momentum of $P$ before the collision.
\item Find, in vector form, the impulse that the wall exerts on $P$.
\item State, in vector form, the impulse that $P$ exerts on the wall.
\item Find in either order.
\begin{itemize}
\item The magnitude of the impulse that the wall exerts on $P$.
\item The angle between $\mathbf { i }$ and the impulse that the wall exerts on $P$.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2024 Q1 [8]}}