## Question 5:

**Part 5(a):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Using horizontal motion (whole motion): $U\cos\alpha \times t = 120$ OR (half way): $U\cos\alpha \times t = 60$ | M1, A1 | 3.3, 1.1b |
| Using vertical motion: $U\sin\alpha \times t - \frac{1}{2}gt^2 = 0$ OR $0 = U\sin\alpha - gt$ | M1, A1 | 3.4, 1.1b |
| Attempt to solve problem by eliminating $t$ | DM1 | 3.1b |
| $U^2 \sin\alpha\cos\alpha = 588$ | A1* | 2.2a |

**Notes 5a:**
- N.B. Could score 2/6 for any one of the 4 given equations if there is no corresponding second equation or the attempt is incorrect
- M1: Complete method to give equation in $U$, $\alpha$ and $t$ only, condone sin/cos confusion and sign errors, each term that needs to be resolved must be resolved
- A1: Correct equation
- DM1: Eliminate $t$, dependent on first and second M1s
- A1*: Given answer correctly obtained, with no wrong working seen. Allow $588 = U^2\sin\alpha\cos\alpha$ but nothing else

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**Part 5(b):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Using vertical motion OR conservation of energy | M1 | 3.4 |
| $0^2 = (U\sin\alpha)^2 - 2g\times10$ OR $\frac{1}{2}mU^2 - \frac{1}{2}m(U\cos\alpha)^2 = mg\times10$ | A1 | 1.1b |
| **Alternative 1:** If $t$ is time to top: use of $10 = \frac{1}{2}gt^2$ oe ($t = \frac{10}{7}$) to obtain equation in $U$ and $\alpha$ only; $U\sin\alpha = 14$ or $U\cos\alpha = 42$ | M1, A1 | |
| **Alternative 2:** use of $10 = U\sin\alpha\cdot t - \frac{1}{2}gt^2$ with $t = \frac{60}{U\cos\alpha}$ substituted; $10 = U\sin\alpha \times \frac{60}{U\cos\alpha} - \frac{1}{2}g\left(\frac{60}{U\cos\alpha}\right)^2$ | M1, A1 | |
| Attempt to solve by eliminating $\alpha$: e.g. $U\sin\alpha = 14 \Rightarrow U\cos\alpha = 42$, square and add; OR $U^2\sin^2\alpha = 20g = 196$ and $U^2\sin\alpha\cos\alpha = 588$, divide to give $\tan\alpha = \frac{1}{3}$ then $\sin^2\alpha = \frac{1}{10}$ | DM1 | 3.1b |
| $U^2 = 1960$ | A1* | 2.2a |

**Notes 5b:**
- N.B. Just stating $\sin^2\alpha = \frac{1}{10}$ with no working is DM0A0
- N.B. Verification (starting with $U^2 = 1960$ and working backwards) is not acceptable

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**Part 5(c):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $V$ less than calculated, since air resistance has to be overcome, or just 'because of air resistance' isw | B1 | 3.5a |

**Notes 5c:**
- B1: Clear statement isw

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**Part 5(d):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. wind effects, more accurate value of $g$, spin of ball, size of ball, shape of ball, dimensions of ball, not a particle, variable acceleration, surface area of ball, humidity. Allow wind resistance and rotational resistance. (Ignore any mention of air resistance or drag) | B1 | 3.5c |

**Notes 5d:**
- B1: B0 if there is an incorrect extra e.g. mass or weight