| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions 1 |
| Type | Vertical drop and bounce |
| Difficulty | Moderate -0.3 This is a straightforward application of coefficient of restitution in a standard bouncing ball scenario. Part (i) requires finding e = ā(122.5/160) ā 0.875 and applying it again; part (ii) involves solving a geometric sequence inequality; part (iii) tests conceptual understanding of e=0 and e=1. While it requires multiple steps and understanding of restitution, the mathematical techniques are routine (geometric sequences, basic algebra) with no novel problem-solving insight needed, making it slightly easier than average. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (i) | Use of š£2 = š¢2+2šš to find v or w |
| Answer | Marks |
|---|---|
| Height = 93.8 (cm) or 0.938 m | M1 |
| Answer | Marks |
|---|---|
| [8] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | Or 1 šš£2 = ššĆ1.6 |
| Answer | Marks |
|---|---|
| 0.93789 | v is arrival speed |
| Answer | Marks |
|---|---|
| (ii) | To get height < 0.1m speed must be < 1.4 m sā1 |
| Answer | Marks |
|---|---|
| n = 11 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.4 |
| Answer | Marks |
|---|---|
| 2.2a | š£2 = 2šĆ0.1 |
| Answer | Marks |
|---|---|
| 8 | Allow = or any inequality; do |
| Answer | Marks | Guidance |
|---|---|---|
| (iii) | (A) | The ball will stay on the horizontal surface after |
| reaching it | B1 | |
| [1] | 2.2a | |
| (B) | The ball will return to a height of 160 cm | |
| (repeatedly) | B1 | |
| [1] | 2.2a | OR bounce off floor at same speed |
Question 5:
5 | (i) | Use of š£2 = š¢2+2šš to find v or w
š£ = 5.6 (m sā1)
š¤ = 4 .9 (m sā1)
š¤ 4.9
Coefficient of restitution is (= )
š£ 5.6
7
= or 0.875
8
Speed before 2nd bounce is also w
Speed after 2nd bounce is šĆš¤ (= 0.875Ć4.9)
Height = 93.8 (cm) or 0.938 m | M1
A1
A1
M1
A1
B1
M1
A1
[8] | 3.1b
1.1
1.1
1.2
1.1
3.3
1.1
1.1 | Or 1 šš£2 = ššĆ1.6
2
š£2 = 02+2šĆ1.225
02 = š¤2ā2šĆ1.225
Soi
343
4.2875 or
80
0.93789 | v is arrival speed
w is leaving speed
122.5= 160Ćš2 M1A1A1
7
š = M1A1
8
2
7
ā = 122.5Ć( ) M1A1A1
8
May see 0.765625 (š2)
(ii) | To get height < 0.1m speed must be < 1.4 m sā1
š
7
Solve 1.4= ( ) Ć5.6 (or 4.9)
8
n = 11 | M1
M1
A1
[3] | 3.4
1.1
2.2a | š£2 = 2šĆ0.1
2š
7
OR 0.1 = ( ) Ć1.6 (or 1.225)
8 | Allow = or any inequality; do
not allow use of 0.099 oe.
Allow for step-by step approach
(iii) | (A) | The ball will stay on the horizontal surface after
reaching it | B1
[1] | 2.2a
(B) | The ball will return to a height of 160 cm
(repeatedly) | B1
[1] | 2.2a | OR bounce off floor at same speed
as it hit the floor5 A small ball is held at a height of 160 cm above a horizontal surface. The ball is released from rest and rebounds from the surface. After its first bounce on the surface the ball reaches a height of 122.5 cm .
\begin{enumerate}[label=(\roman*)]
\item Find the height reached by the ball after its second bounce on the surface.
After $n$ bounces the height reached by the ball is less than 10 cm .
\item Find the minimum possible value of $n$.
\item State what would happen if the same ball is released from rest from a height of 160 cm above a different horizontal surface and\\
(A) the coefficient of restitution between the ball and the new surface is 0 ,\\
(B) the coefficient of restitution between the ball and the new surface is 1 .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2018 Q5 [13]}}