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2021 Paper 2 Q10
D: 1500.0 B: 1500.0
non-inertial frame constrained motion energy method bead on wire constant acceleration parabolic constraint lagrangian mechanics

A train moves westwards on a straight horizontal track with constant acceleration \(a\), where \(a > 0\). Axes are chosen as follows: the origin is fixed in the train; the \(x\)-axis is in the direction of the track with the positive \(x\)-axis pointing to the East; and the positive \(y\)-axis points vertically upwards. A smooth wire is fixed in the train. It lies in the \(x\)--\(y\) plane and is bent in the shape given by \(ky = x^2\), where \(k\) is a positive constant. A small bead is threaded onto the wire. Initially, the bead is held at the origin. It is then released.

  1. Explain why the bead cannot remain stationary relative to the train at the origin.
  2. Show that, in the subsequent motion, the coordinates \((x, y)\) of the bead satisfy \[ \dot{x}(\ddot{x} - a) + \dot{y}(\ddot{y} + g) = 0 \] and deduce that \(\tfrac{1}{2}(\dot{x}^2 + \dot{y}^2) - ax + gy\) is constant during the motion.
  3. Find an expression for the maximum vertical displacement, \(b\), of the bead from its initial position in terms of \(a\), \(k\) and \(g\).
  4. Find the value of \(x\) for which the speed of the bead relative to the train is greatest and give this maximum speed in terms of \(a\), \(k\) and \(g\).

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2015 Paper 2 Q11
D: 1600.0 B: 1484.0
circular motion inextensible string impulse conservation of momentum conservation of energy elastic collision constrained motion particle system kinematics

Three particles, \(A\), \(B\) and \(C\), each of mass \(m\), lie on a smooth horizontal table. Particles \(A\) and \(C\) are attached to the two ends of a light inextensible string of length \(2a\) and particle \(B\) is attached to the midpoint of the string. Initially, \(A\), \(B\) and \(C\) are at rest at points \((0,a)\), \((0,0)\) and \((0,-a)\), respectively. An impulse is delivered to \(B\), imparting to it a speed \(u\) in the positive \(x\) direction. The string remains taut throughout the subsequent motion.

TikZ diagram
  1. At time \(t\), the angle between the \(x\)-axis and the string joining \(A\) and \(B\) is \(\theta\), as shown in the diagram, and \(B\) is at \((x,0)\). Write down the coordinates of \(A\) in terms of \(x,a\) and \(\theta\). Given that the velocity of \(B\) is \((v,0)\), show that the velocity of \(A\) is \((\dot x + a\sin\theta \,\dot \theta\,,\, a\cos\theta\, \dot\theta)\), where the dot denotes differentiation with respect to time.
  2. Show that, before particles \(A\) and \(C\) first collide, \[ 3\dot x + 2a \dot\theta \sin\theta =u \text{ and } \dot \theta^2 = \frac{u^2}{a^2(3-2\sin^2\theta)} \,. \]
  3. When \(A\) and \(C\) collide, the collision is elastic (no energy is lost). At what value of \(\theta\) does the second collision between particles \(A\) and \(C\) occur? (You should justify your answer.)
  4. When \(v=0\), what are the possible values of \(\theta\)? Is \(v =0\) whenever \(\theta\) takes these values?

Solution:

  1. \(A\) has coordinates \((x-a\cos \theta, a\sin \theta)\). Differentiating with respect to \(t\) the velocity of \(A\) is \((\dot{x}+a\sin \theta \cdot \dot{\theta}, a \cos \theta \cdot \dot{\theta})\)
  2. By considervation of momentum \(\rightarrow\) we must have \(mu = m(\dot{x}+a\dot{\theta}\sin \theta) + m\dot{x} + m(\dot{x}+a\dot{\theta}\sin \theta) = m(3\dot{x} + 2a \dot{\theta} \sin \theta)\) and the first equation follows. By conservation of energy, we must have \begin{align*} && \frac12 m u^2 &= \frac12 m \dot{x}^2 + \frac12m((\dot{x}+a\dot{\theta}\sin \theta)^2 + a^2 \dot{\theta}^2 \cos^2\theta ) + \frac12m((\dot{x}+a\dot{\theta}\sin \theta)^2 + a^2 \dot{\theta}^2 \cos^2\theta ) \\ &&&= \frac32m\dot{x}^2 + 2m a\dot{x}\dot{\theta}\sin \theta + ma^2\dot{\theta}^2(\sin^2\theta+\cos^2\theta) \\ \Rightarrow && u^2 &= \dot{x}(3\dot{x} + 4a \dot{\theta} \sin \theta) + 2a^2\dot{\theta}^2 \\ &&&= \left ( \frac{u-2a\dot{\theta}\sin \theta}{3}\right)\left ( 3\left ( \frac{u-2a\dot{\theta}\sin \theta}{3}\right)+ 4a \dot{x}\dot{\theta} \sin \theta \right) + 2a^2\dot{\theta}^2 \\ \Rightarrow && 3u^2 &= (u - 2a\dot{\theta} \sin \theta)^2 + 4a(u - 2 a \dot{\theta} \sin \theta) \dot{\theta}\sin \theta + 6a^2 \dot{\theta}^2 \\ &&&= u^2 + 4a^2\dot{\theta}^2 \sin^2 \theta - 8a^2\dot{\theta}^2\sin^2\theta + 6a^2 \dot{\theta}^2 \\ \Rightarrow && \dot{\theta}^2 &= \frac{u^2}{a^2(3-2\sin^2\theta)} \end{align*}
  3. Since \(\dot{\theta}^2 > 0\) \(\theta\) is strictly increasing or decreasing, therefore the first collision will be when \(\theta = 0\), the second when \(\theta = \pi\)
  4. If \(v = 0\), from our first equation we have \(2a \dot{\theta} \sin \theta = u \Rightarrow \dot{\theta}^2 = \frac{u^2}{4a^2 \sin^2 \theta} = \frac{u^2}{a^2(3-2\sin^2\theta)}\) so \(4\sin^2 \theta = 3 - 2\sin^2 \theta \Rightarrow \sin^2 \theta = \frac{1}{2}\) therefore the angles are all the multiples of \(\frac{\pi}{4}\).

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