Standard +0.3 This is a straightforward mechanics problem requiring application of SUVAT equations to two particles under gravity. Students must set up equations for both particles (s = ut + Β½gtΒ²), use the constraint that both land at t=5, and perform basic algebraic manipulation including factoring and square roots. While it requires careful setup and the 'prove that' format demands clear justification, the mathematical techniques are standard A-level mechanics with no novel insight requiredβslightly easier than average.
Two particles \(A\) and \(B\) are released from rest from different starting points above a horizontal surface.
\(A\) is released from a height of \(h\) metres.
\(B\) is released at a time \(t\) seconds after \(A\) from a height of \(kh\) metres, where \(0 < k < 1\)
Both particles land on the surface \(5\) seconds after \(A\) was released.
Assuming any resistance forces may be ignored, prove that
$$t = 5(1 - \sqrt{k})$$
Fully justify your answer.
[5 marks]
Question 16:
16 | 1
Uses s =ut+ at2 for particle A
2
With u=0 t=5 and a=g
(g can be 9.8 9.81, or 10) | 3.1b | M1 | 1 2
π π = π’π’π‘π‘+ πππ‘π‘
, 2 and
π π = β,ππ = π¨π¨ π’π’ = 0 π‘π‘ = 5
25
β = ππ2
1 2
ππβ = π¨π¨(5βπ‘π‘)
2
25 1 2
π¨π¨ππ = π¨π¨(5βπ‘π‘)
2 2
, since 0 < t <5
5βππ = 5βπ‘π‘
π‘π‘ = 5οΏ½1ββπποΏ½
Finds in terms of
Condone use of g, 9.8, 9.81, or 10
Can acβhieve this mππ ark for
h=12.5g,122.5,122.625,125 | 1.1b | A1
1
Uses kh=ut+ at2 for particle B
2
With u=0 and a=g
(g can be 9.8 9.81, or 10) | 1.1a | M1
Deduces βtime for Bβ + t =5 | 3.3 | B1
Eliminates h and completes
reasoned argument to obtain
with consistent use of g or value for
gπ‘π‘ a=nd5 οΏ½n1oβ slβipππsοΏ½
AG | 2.1 | R1
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution
Two particles $A$ and $B$ are released from rest from different starting points above a horizontal surface.
$A$ is released from a height of $h$ metres.
$B$ is released at a time $t$ seconds after $A$ from a height of $kh$ metres, where $0 < k < 1$
Both particles land on the surface $5$ seconds after $A$ was released.
Assuming any resistance forces may be ignored, prove that
$$t = 5(1 - \sqrt{k})$$
Fully justify your answer.
[5 marks]
\hfill \mbox{\textit{AQA Paper 2 2020 Q16 [5]}}