The correct answer is C because:

Resolve the forces horizontally and apply Newton’s second law:
- P + T₂sin(θ) − T₁sin(θ) = ma_x = 0
- P = T₁sin(θ) − T₂sin(θ)
- P = (T₁ - T₂ )sin(θ) [Equation 1]

Resolve the forces vertically and apply Newton’s second law:
- T₁cos(θ) + T₂cos(θ) − mg = ma_y = 0
- mg = (T₁ + T₂) cos(θ) [Equation 2]
Obtain the ratio of the two equations:
- $\frac{[E q u a t i o n 1]}{[E q u a t i o n 2]}$
- $\frac{p}{m g} = \frac{(T_{1} - T_{2}) s i n (θ)}{(T_{1} + T_{2}) c o s (θ)} = \frac{(T_{1} - T_{2})}{(T_{1} + T_{2})} t a n (θ)$
- $\frac{p}{m g t a n (θ)} = \frac{(T_{1} - T_{2})}{(T_{1} + T_{2})}$
Rearrange to obtain the ratio :
- $\frac{p}{m g t a n (θ)} T_{1} + \frac{p}{m g t a n (θ)} T_{2} = (T_{1} - T_{2})$
- $T_{1} - \frac{p}{m g t a n (θ)} T_{1} = T_{2} + \frac{p}{m g t a n (θ)} T_{2}$
- $T_{1} (1 - \frac{p}{m g \tan (θ)}) = T_{2} (1 + \frac{p}{m g \tan (θ)})$
- $\frac{T_{1}}{T_{2}} = \frac{(1 + \frac{P}{m g t a n (θ)})}{(1 - \frac{P}{m g t a n (θ)})}$

B is incorrect because sine and cosine are used the wrong way around.

A and D are incorrect because the plus and minus signs have been mixed up.

In this question, be careful that the angle of the ropes from the vertical = 90° - the angle from the horizontal as shown in the diagram below.

2-3-mcq-h-ma5-ocr-al-physics

Taking a ratio of the two equations often results in certain terms cancelling down, like sin(θ) ÷ cos(θ) = tan(θ). This is a good clue for how this question needs to be solved. Go through your algebraic manipulations once more to check you haven’t made this mistake.

	Component of weight parallel to slope / N	Component of weight perpendicular to slope / N
A	2.2	8.2
B	2.5	8.1
C	2.8	8.0
D	3.1	7.9

Scalars & Vectors (OCR A Level Physics)

Exam Questions