### Part (a)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $F = (-)\frac{k}{x^2}$ | | |
| $x = R$ $F = (-)mg$ | | |
| $mg = (-){\frac{k}{R^2}}$ | M1 | Using NL2 at the surface of the Earth with the proportionality condition, force to be $mg$. distance to be $R$. Can use $F = \frac{GMm}{x^2}$ |
| Mag of $F = \frac{mgR^2}{x^2}$ * | A1cso | (2) Find the constant of proportionality and hence deduce the stated magnitude. |

### Part (b)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(m)v\frac{dv}{dx} = -(m)\frac{gR^2}{x^2}$ | M1 | Using NL2 with acceleration in form $v\frac{dv}{dx}$. Minus sign may be missing. $m$ may be cancelled. |
| $\int vdv = -\int gR^2 x^{-2}dx$ | DM1A1ft | Attempt integration of their expression. $v^2$ or $x^{-1}$ must be seen. Correct integration follow through a missing minus sign. Constant of integration may be missing. |
| $\frac{1}{2}v^2 = gR^2 x^{-1} + (c)$ | | |
| $x = R$ $v = U$ $\frac{1}{2}U^2 = gR + c$ | M1 | |
| $\frac{1}{2}v^2 = \frac{gR^2}{x} + \frac{1}{2}U^2 - gR$ oe | A1 | |
| $x = \frac{21}{20}R$ $v = 0$ $0 = \frac{gR^2}{\frac{21}{20}R} + \frac{1}{2}U^2 - gR$ | M1 | Using $x = R$ $v = U$ OR substitute $x = \frac{21}{20}R$ $v = 0$ in an equation containing $c$ |
| $\frac{1}{2}U^2 = gR - \frac{20}{21}gR$ | | |
| $U = \sqrt{\frac{2gR}{21}}$ oe | A1 | (7) |

**Total: [9]**

**Key Notes:**
- (a)M1: Using NL2 at the surface of the Earth with the proportionality condition, force to be $mg$, distance to be $R$. Can use $F = \frac{GMm}{x^2}$
- (a)A1cso: Find the constant of proportionality and hence deduce the stated magnitude.
- (b)M1: Using NL2 with acceleration in form $v\frac{dv}{dx}$. Minus sign may be missing. $m$ may be cancelled.
- (b)DM1: Attempt integration of their expression. $v^2$ or $x^{-1}$ must be seen.
- (b)A1ft: Correct integration follow through a missing minus sign. Constant of integration may be missing.
- (b)M1: Using $x = R$ $v = U$ OR substitute $x = \frac{21}{20}R$ $v = 0$ in an equation containing $c$
- (b)A1: Substituting a correct constant to obtain a correct expression for $\frac{1}{2}v^2$
- (b)M1: Substitute $x = \frac{21}{20}R$ $v = 0$ OR substitute $x = R$ $v = U$
- (b)A1: Complete to a correct expression for $U$.

**Definite integration in (b):**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(m)v\frac{dv}{dx} = -(m)\frac{gR^2}{x^2}$ | M1 | |
| $\int_U^0 vdv = -\int_R^{\frac{21R}{20}} gR^2 x^{-2}dx$ | | |
| $\left[\frac{1}{2}v^2\right] = \left[gR^2 x^{-1}\right]$ | DM1A1ft | (ft missing sign, as main scheme) Limits not needed |
| $\left[\frac{1}{2}v^2\right]_U^0 = \left[gR^2 x^{-1}\right]_R^{\frac{21R}{20}}$ | M1 | limits seen, correct "values" but may be paired incorrectly |
| | A1 | correct limits, $0$ and $\frac{21R}{20}$ at top or bottom on both integrals, $U$ and $R$ in the other positions |
| M1 | Substitute their limits | A1 | correct answer. |

**NB:** It is possible to appear to omit the minus sign and reverse the limits on one integral. However, without some explanation this will be insufficient working and scores as a missing minus case.

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