2003 Paper 3 Q11

Year: 2003
Paper: 3
Question Number: 11

Course: LFM Pure and Mechanics
Section: Constant Acceleration

Difficulty: 1700.0 Banger: 1486.9

constant acceleration relative motion collision inequality trigonometry projectile bearing optimisation

Problem

Point \(B\) is a distance \(d\) due south of point \(A\) on a horizontal plane. Particle \(P\) is at rest at \(B\) at \(t=0\), when it begins to move with constant acceleration \(a\) in a straight line with fixed bearing~\(\beta\,\). Particle \(Q\) is projected from point \(A\) at \(t=0\) and moves in a straight line with constant speed \(v\,\). Show that if the direction of projection of \(Q\) can be chosen so that \(Q\) strikes \(P\), then \[ v^2 \ge ad \l 1 - \cos \beta \r\;. \] Show further that if \(v^2 >ad(1-\cos\beta)\) then the direction of projection of \(Q\) can be chosen so that \(Q\) strikes \(P\) before \(P\) has moved a distance \(d\,\).

No solution available for this problem.

Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1486.9

Banger Comparisons: 1

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Problem source

Point $B$ is a distance $d$ due south of point $A$ on a horizontal plane. 
Particle $P$ is at rest at $B$ at $t=0$, when it begins 
to move with constant acceleration $a$ in a straight line
with  fixed bearing~$\beta\,$. 
Particle $Q$ is projected  from 
point $A$ at $t=0$ and moves in a straight line with constant
speed $v\,$. Show that if the direction of projection of $Q$ can be chosen so that
$Q$ strikes $P$, then   
\[   
v^2 \ge ad \l 1 - \cos \beta \r\;.   
\]   
Show further that if $v^2 >ad(1-\cos\beta)$ then the direction of projection of $Q$
can be chosen so that $Q$ strikes $P$ before $P$ has moved a distance $d\,$.

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