| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Successive collisions with wall rebound |
| Difficulty | Challenging +1.2 This is a standard two-stage collision problem requiring conservation of momentum, Newton's restitution law, and inequality reasoning to prevent further collisions. Part (a) is routine algebraic manipulation to show direction reversal. Part (b) requires setting up inequalities based on relative velocities after both collisions, which is a common Further Mechanics technique but requires careful bookkeeping across multiple collision stages. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | m20 + m–10 = mv + mv |
| Answer | Marks |
|---|---|
| motion of B is reversed). | M1* |
| Answer | Marks |
|---|---|
| [5] | 3.3 |
| Answer | Marks |
|---|---|
| 2.2a | DR |
| Answer | Marks |
|---|---|
| u_b and v_b (may see 0 e ) | If v is reversed |
Question 6:
6 | (a) | m20 + m–10 = mv + mv
A B
v − v
B A1 = e
2 0 − − 0
m20 + m–10 = mv + mv
A B
v − v
B A1 = e
2 0 − − 0
2v = 10 + 30e
B
v = 5 + 15e > 0 while u = -10 < 0
B B
or v = 5 + 15e which is positive while
B
original velocity of B was negative (so the
velocity changes sign and hence direction of
motion of B is reversed). | M1*
M1*
A1
M1dep
*
A1
[5] | 3.3
3.3
1.1
1.1
2.2a | DR
Attempt at conservation of
momentum
Attempt at NEL
Both equations correct and
consistent
v + v = 10
B A
v – v = 30e
B A
Combining two equations of
correct form using simultaneous
equations to find v
B
Need to see clear comparison of
u_b and v_b (may see 0 e ) | If v is reversed
A
m20 + m–10 = -mv + mv
A B
v + v
B A1 = e
2 0 − − 0
Both equations correct and
consistent
v - v = 10
B A
v + v = 30e
B A
2v = 10 + 30e
B6 Two identical spheres, $A$ and $B$, each of mass $m \mathrm {~kg}$, are moving directly towards each other along the same straight line on a smooth horizontal surface until they collide. Just before they collide, the speeds of $A$ and $B$ are $20 \mathrm {~ms} ^ { - 1 }$ and $10 \mathrm {~ms} ^ { - 1 }$ respectively. The coefficient of restitution between $A$ and $B$ is $e$.
\begin{enumerate}[label=(\alph*)]
\item By finding, in terms of $e$, an expression for the velocity of $B$ after the collision, show that the direction of motion of $B$ is reversed by the collision.
After the collision between $A$ and $B$, which is not perfectly elastic, $B$ goes on to collide directly with a fixed, vertical wall. The coefficient of restitution between $B$ and the wall is $\frac { 2 } { 5 } e$. After the collision between $B$ and the wall, there are no further collisions between $A$ and $B$.
\item Determine the range of possible values of $e$.\\
$7 \quad$ A body $B$ of mass 1.5 kg is moving along the $x$-axis. At the instant that it is at the origin, $O$, its velocity is $u \mathrm {~ms} ^ { - 1 }$ in the positive $x$-direction.
At any instant, the resistance to the motion of $B$ is modelled as being directly proportional to $v ^ { 2 }$ where $v \mathrm {~ms} ^ { - 1 }$ is the velocity of $B$ at that instant. The resistance to motion is the only horizontal force acting on $B$.
At an instant when $B$ 's velocity is $2 \mathrm {~ms} ^ { - 1 }$, the resistance to its motion is 24 N .
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2024 Q6 [12]}}