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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2025

Paper 1

Since $y_1=(Q_s+Q_a)/2$, $y_3=(Q_s-Q_a)/2$, elimination of the middle coordinate by the center of mass gives
$$
y_2=\frac{TY-mQ_s}{M}.
$$
After substitution,
$$
\widehat L=\frac m4(\dot Q_s^2+\dot Q_a^2)
+\frac{(T\dot Y-m\dot Q_s)^2}{2M}
-\frac{kT^2}{M^2}\left(Y-\frac{Q_s}{2}\right)^2
-\frac k4Q_a^2-TgY.
$$
The antisymmetric coordinate is decoupled. Its Euler--Lagrange equation is
$$
\boxed{\frac m2\ddot Q_a+\frac k2Q_a=0,\qquad\text{or}\qquad
\ddot Q_a+\frac kmQ_a=0.}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 8B / b / Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 8B / b / Solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 8B / b / Solution

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