# Question 6

## 6(b)

**X component:**

M1 | Allow with $F$. Requires all terms - condone trig confusion

$$\frac{5mga}{b} = F\sin\theta\cos\theta + \sin\theta$$

A1 | $F$ substituted

$$\frac{5mga}{b}$$

**Y component:**

M1 | Allow with $F$. Requires all terms - condone trig confusion and sign errors

$$\frac{5mga}{4mg} = F\cos^2\theta - 4mg\cos\theta$$

A1 | Correct unsimplified

A1 | Correct substituted

---

## 6(c)

M1 | 

$$\tan\theta = \frac{\tan\theta_Y}{\tan\theta_X} = \frac{\frac{5a}{4}\cos^2\theta}{b}$$

$$\frac{5a}{b}\cos\sin\theta$$

$$\frac{5a}{4}\cos^2\theta - \frac{5a}{b}\sin^2\theta$$

$$\frac{5a}{4b} - \frac{5a}{b} \cdot \frac{5}{4}$$

A1 | CSO

---

## 7a

M1 | Equate horizontal components of speeds:

$$u\cos\theta = 6\cos 45° = 3\sqrt{2} = 4.24\ldots$$

A1 | Correct unsimplified

M1 | Use suvat for vertical speeds: condone sign errors

$$u\sin\theta = 2g - 6\sin 45° = 2g - 3\sqrt{2}$$

A1 | Correct unsimplified

DM1 | Dependent on previous 2 Ms. Follow their components.

$$\tan\theta = \frac{2g - 3\sqrt{2}}{3\sqrt{2}} = 3.61\ldots \approx 74.6° \text{ or } 75°$$

A1 | $u = 15.93\ldots$ (m s$^{-1}$)

---

## 7b

B1 | At max height, speed $= u\cos\theta = 3\sqrt{2}$ (m s$^{-1}$)

M1 | Correct for their $v$ at the top, $v \neq 0$

$$KE = \frac{1}{2} \times 0.7 \times (3\sqrt{2})^2$$

A1 | Accept awrt 6.30. CSO

$$= 6.3 \text{ (J)}$$

---

## 7c

M1 | Use suvat to find first time $v = 6$

$$u\sin\theta - t = 3\sqrt{2}$$

A1 | 

$$2g - 3\sqrt{2} - gt = 3\sqrt{2}$$

M1 | Solve for $t$

$$t = \frac{2g - 6\sqrt{2}}{g} = 1.13\ldots$$

A1 | Sensitive to premature approximation. Allow 1.14.

A1 | CAO accept awrt 0.87

$$T = 2 \times 1.13 - 0.87$$

---

## Alt 7c (Method 1)

M1A1 | Find time from top to A:

$$6\sin 45° - gt = \frac{12\sqrt{2}}{2}$$

M1 | 

$$T = 2t - \frac{g}{0.87}$$

A1 | Correct strategy

A1 | Correct unsimplified

---

## Alt 7c (Method 2)

M1A1 | Time to top: $u\sin\theta - gt$ (their $u$)

$$t = 1.567\ldots$$

M1A1 | 

$$T = 2 \times 2 - 1.567\ldots = 0.87$$

A1 | 

---

## Alt 7c (Method 3)

M1 | Vertical speed at A $= -$ (vertical speed at B) $= -\sqrt{36 - (3\sqrt{2})^2 + 3\sqrt{2}}$

A1 | Or use the $45°$ angle

M1 | Use suvat for AB. Correct use for their values

$$3\sqrt{2} - 3\sqrt{2} - gT$$

A1 | 

$$T = 0.87$$

A1 | 

---

## Alt 7d

M1 | Form expression for $v^2$. Inequality not needed at this stage

$$v^2 = (3\sqrt{2})^2 + u\sin\theta t^2 = 36$$

A1 | Correct inequality for $v^2$

$$18 - u\sin\theta t = 18$$

M1 | 

$$t = \frac{u\sin\theta - 18}{g} = \frac{u\sin\theta - 18}{g}$$

A1 | 

$$T = \frac{u\sin\theta - 18}{g} + \frac{u\sin\theta - 18}{g} = \frac{2 \times 18}{g} = 0.866$$

A1 |