Standard +0.3 This is a straightforward dimensional analysis problem requiring students to equate dimensions on both sides of an equation. While it involves multiple terms and careful tracking of dimensions, dimensional analysis is a standard technique taught in mechanics modules, and the algebraic manipulation required is routine. The 4-mark allocation reflects the methodical but unchallenging nature of the work.
\includegraphics{figure_2}
A particle is projected with speed \(v\) from a point O on horizontal ground. The angle of projection is \(\theta\) above the horizontal. The particle passes, in succession, through two points A and B, each at a height \(h\) above the ground and a distance \(d\) apart, as shown in the diagram.
You are given that \(d^2 = \frac{v^\alpha \sin^2 2\theta}{g^\beta} - \frac{8h^2 \cos^2 \theta}{g}\).
Use dimensional analysis to find \(\alpha\) and \(\beta\). [4]
Question 2:
2 | [ v ]=LT −1 or [ g ]=LT −2
( −1 )α ( −1 )α ( −1 )2
LT LT L LT
L2 = or =
( −2 )β ( −2 )β LT −2
LT LT
α−β=2
−α+2β=0
α=4,β=2 | B1
M1*
M1dep*
A1
[4] | 1.2
2.1
1.1a
1.1 | Correctly stating the dimensions of either v or g
Setting up an equation in L and T using either
v α sin22θ v α sin22θ 8hv2cos2θ
d2= or =
g β g β g
Setting up two consistent equations in αand β
\includegraphics{figure_2}
A particle is projected with speed $v$ from a point O on horizontal ground. The angle of projection is $\theta$ above the horizontal. The particle passes, in succession, through two points A and B, each at a height $h$ above the ground and a distance $d$ apart, as shown in the diagram.
You are given that $d^2 = \frac{v^\alpha \sin^2 2\theta}{g^\beta} - \frac{8h^2 \cos^2 \theta}{g}$.
Use dimensional analysis to find $\alpha$ and $\beta$. [4]
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2022 Q2 [4]}}