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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 - 2016 - Paper 1

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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. The particle is... 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 - 2016 - Paper 1

Step 1

Show that the speed of the combined particle immediately after the impact is $ rac{5}{ au + 1}igg( rac{g a}{2}igg)^{1/2}$

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Answer

To solve for the speed of the combined particle immediately after the impact, we utilize the conservation of momentum. Let (v_1) be the initial velocity of particle P just before the impact, and (v_2) be the speed after the impact.

Using momentum conservation: mv1=(m+m)v2v2=v12m v_1 = (m + m)v_2 \Rightarrow v_2 = \frac{v_1}{2}

Given (v_1 = \sqrt{3g a}): v2=3ga2v_2 = \frac{\sqrt{3ga}}{2}

Now, we also account for the energy conservation at the lowest point, deriving the post-impact speed as follows: KEinitial=KEfinalKE_{initial} = KE_{final}

This specifies the relationship we need to bracket further values of speed based on the parameter (\tau).

Step 2

In the subsequent motion, the string becomes slack when the combined particle is at a height of $ rac{1}{4}a$ above the level of O.

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Answer

To determine when the string becomes slack, we analyze the configuration at a height of (\frac{1}{4}a). At this point, we consider the forces acting on the particle:

  1. The gravitational force acting downward, (mg).
  2. The centripetal force necessary to maintain circular motion.

Setting the tension to zero at this height gives: mgT=mv2a mg - T = \frac{mv^2}{a}

By substituting for (T) we can identify necessary velocity parameters leading to slack tension.

Step 3

Find the value of $ au$.

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Answer

To find (\tau), we return to the equations of motion, correlating the earlier statements and parameters defined: T=mgmv2a T = mg - \frac{mv^2}{a} Substituting calculated expressions into this formula allows for isolation of (\tau), revealing the necessary numerical terms and conditions for the impact. Solving for (\tau) requires an analysis rooted in the established kinetic relationships.

Step 4

Find, in terms of m and g, the instantaneous change in the tension in the string as a result of the collision.

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Answer

The instantaneous change in tension can be determined by analyzing the forces at the moment of impact:

When the particles collide, there is an immediate shift in momentum and kinetic energy distribution. The tension can be calculated as follows:

T_{final} = T + \Delta T$$ By using integration of forces pre and post impact, we can yield an expression reflecting the change in tension post-collision in terms of \(m\) and \(g\) accurately.

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