| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Vertical drop and bounce |
| Difficulty | Challenging +1.8 This is a challenging Further Mechanics question requiring understanding of coefficient of restitution, infinite geometric series for bouncing, and impulse-momentum. Part (a) demands setting up the total time equation involving the sum of an infinite GP (time = √(2h/g)(1+e)/(1-e)), which requires significant insight beyond routine mechanics. Part (b) is more straightforward application of impulse = m(v_after - v_before). The 8-mark allocation and need for extended reasoning with infinite series places this well above average difficulty. |
| Spec | 6.03f Impulse-momentum: relation6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | 78.4= 1(9.8)t2(⇒t =4) |
| 2 1 1 | B1 | 3.3 |
| Answer | Marks |
|---|---|
| release to first bounce | Or correct speed at the |
| Answer | Marks | Guidance |
|---|---|---|
| 1 1 | M1* | 3.4 |
| Answer | Marks |
|---|---|
| 1 | oe suvat equation |
| Answer | Marks | Guidance |
|---|---|---|
| 2 1 | A1ft | 1.1 |
| Answer | Marks |
|---|---|
| 1 | v =39.2e |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1* | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | A1 | 1.1 |
| 4+8e+8e2 +...=6 | M1dep* | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1−e | M1 | 3.1a |
| Answer | Marks |
|---|---|
| 0 1 2 | Dependent on all |
| Answer | Marks | Guidance |
|---|---|---|
| 8e=2(1−e)⇒e=0.2 | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (b) | 23.52 = m ( get −(−gt )) |
| 1 1 | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 9.8(0.2)(4)+9.8(4) | A1 | 1.1 |
Question 6:
6 | (a) | 78.4= 1(9.8)t2(⇒t =4)
2 1 1 | B1 | 3.3 | Correct equation for the time t from
1
release to first bounce | Or correct speed at the
first impact for B1M1
v =gt
1 1 | M1* | 3.4 | Use of v=u+at with u = 0 to find the
speed v at the first impact
1 | oe suvat equation
(v =39.2)
1
v =get
2 1 | A1ft | 1.1 | Correct expression for the speed v after
2
first impact in terms of their t
1 | v =39.2e
2
1
0=(get )t − gt2
1 2 2
2 | M1* | 3.4 | Use of s=ut+0.5at2 with s = 0 to find
the time between first and second impact
t
2
t =8e
2 | A1 | 1.1
4+8e+8e2 +...=6 | M1dep* | 2.1 | Setting up an infinite series (at least three
terms on the lhs) equal to 6
1
4+8e(1+e+...)=6⇒4+8e =6
1−e | M1 | 3.1a | Correct use of the sum to infinity formula
for a GP of the form u +eu +e2u +...
0 1 2 | Dependent on all
previous M marks
8e=2(1−e)⇒e=0.2 | A1 | 2.2a
[8]
6 | (b) | 23.52 = m ( get −(−gt ))
1 1 | M1 | 3.3 | Use of impulse = change in momentum | Allow with no value
of e substituted
23.52
m= =0.5
9.8(0.2)(4)+9.8(4) | A1 | 1.1
[2]A small ball of mass $m$ kg is held at a height of $78.4$ m above horizontal ground. The ball is released from rest, falls vertically and rebounds from the ground. The coefficient of restitution between the ball and ground is $e$.
The ball continues to bounce until it comes to rest after $6$ seconds.
\begin{enumerate}[label=(\alph*)]
\item Determine the value of $e$. [8]
\item Given that the magnitude of the impulse that the ground exerts on the ball at the first bounce is $23.52$ Ns, determine the value of $m$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2020 Q6 [10]}}