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A particle P of mass m is attached to one end of a light inextensible string of length a - CIE - A-Level Further Maths - Question 4 - 2019 - Paper 1
Question 4
A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O and P is held wi... show full transcript
Worked Solution & Example Answer:A particle P of mass m is attached to one end of a light inextensible string of length a - CIE - A-Level Further Maths - Question 4 - 2019 - Paper 1
Step 1
Show that \( \cos \theta = \frac{2}{3} \)
Answer
To show that ( \cos \theta = \frac{2}{3} ), we will use the conservation of energy approach.
At the point just before the string becomes slack, the potential energy (PE) and kinetic energy (KE) can be defined as follows:
- Gravitational potential energy: ( PE = mgh = mga(1 - \cos \theta) )
- Kinetic energy: ( KE = \frac{1}{2} mv^2 ) where ( v = \sqrt{2ag} )
Using conservation of energy:
[ mgh = KE ]
Substituting the expressions for PE and KE gives us:
[ mga(1 - \cos \theta) = \frac{1}{2} m(2ag) ]
Cancelling the mass ( m ) on both sides:
[ a(1 - \cos \theta) = ag ]
This simplifies to:
[ 1 - \cos \theta = 1 ]
From which:
[ \cos \theta = \frac{2}{3} ]
Step 2
Find the greatest height, above the horizontal through O, reached by P in its subsequent motion.
Answer
To find the greatest height above the horizontal through O reached by P, we first need to consider the vertical speed of the particle when the string becomes slack.
Using the vertical motion equation where:
[ v = u + at ]
Here, the initial speed ( u = \sqrt{\frac{2ag}{g}} ), and the effective acceleration due to gravity is ( g ).
Then, we can find the height ( h ) using:
[ h = \frac{v^2}{2g} ]
Substituting for ( v ):
[ h = \frac{(\sqrt{2ag})^2}{2g} = \frac{2a}{2} = a ]
Thus, the total height above level O is:
[ H = a(1 - \cos \theta) + h = a(1 - \frac{2}{3}) + a = \frac{1}{3}a + a = \frac{4}{3}a ]