| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Session | Specimen |
| Marks | 10 |
| Topic | Momentum and Collisions 1 |
| Type | Vertical drop and bounce |
| Difficulty | Standard +0.3 This is a straightforward application of Newton's law of restitution and kinematic equations. Part (i) requires standard derivations using v²=u²+2as and v=u+at; part (ii) is simple substitution; part (iii) involves calculating an arithmetic mean and stating that air resistance causes the discrepancy; part (iv) requires recognizing that lower gravity increases time but doesn't affect the coefficient of restitution. All steps are routine mechanics with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae6.03l Newton's law: oblique impacts |
**(i)** Initial impact speed $V = \sqrt{2gh}$ (using $v^2 = u^2 + 2as$) B1
Rebound speed $u = eV$ B1
Substituting into at least one equation for rebound phase: M1
$0^2 = (eV)^2 + 2(-g)H$
$0 = eV + (-g)T$ A1
$H = e^2h$ and $T = e\sqrt{\frac{2h}{g}}$ **AG** **[4]**
**(ii)** $(e = 0.5)$ and $T = 0.3$ s (using $h = 1.8$ m and $g = 10$) B1 **[1]**
**(iii)** The mean of the rebound heights is
$\frac{1}{n}\sum_1^n kH = \frac{H}{n} \times \frac{1}{2}n(n+1)$ M1
$= \frac{1}{2}H(n+1)$ A1
$3.6$ m (using $n = 15$, $H = 0.45$ m) A1
Either the effect of air resistance or a deviation from Newton's law for high impact speeds. B1 **[4]**
**(iv)** Only the time depends on the local gravitational constant. B1 **[1]**13 Professor Oldham wishes to illustrate and test Newton's experimental law of impacts. A ball is dropped from rest from a height $h$ above a rigid horizontal board and rebounds to a height $H$. The time taken to reach the height $H$ after the first impact is $T$. These quantities are recorded using very accurate measuring devices.\\
(i) Show that
$$H = e ^ { 2 } h \quad \text { and } \quad T = e \sqrt { \frac { 2 h } { g } }$$
are predicted by Newton's law, where $e$ is the coefficient of restitution between the ball and the board.\\
(ii) If $h = 180 \mathrm {~cm}$ and $H = 45 \mathrm {~cm}$, determine $T$ from these formulae.
The experiment is repeated for initial heights $h , 2 h , 3 h , \ldots , 15 h$ where $h = 180 \mathrm {~cm}$. The corresponding rebound heights and times taken to reach that height after the first impact are recorded. The mean of the 15 rebound heights is found to be 3.3 m .\\
(iii) Find the mean of the rebound heights predicted by Newton's law and give one reason why this differs from the experimental value.
Professor Oldham is able to repeat the experiment on the surface of the moon using the same experimental set-up inside a laboratory.\\
(iv) The mean of the rebound heights is unchanged, but the mean of the rebound times is substantially increased. Comment on these findings.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 Q13 [10]}}