Problems

A spaceship of mass \(M\) is at rest. It separates into two parts in an explosion in which the total kinetic energy released is \(E\). Immediately after the explosion the two parts have masses \(m_{1}\) and \(m_{2}\) and speeds \(v_{1}\) and \(v_{2}\) respectively. Show that the minimum possible relative speed \(v_{1}+v_{2}\) of the two parts of the spaceship after the explosion is \((8E/M)^{1/2}.\)

Three small spheres of masses \(m_{1},m_{2}\) and \(m_{3},\) move in a straight line on a smooth horizontal table. (Their order on the straight line is the order given.) The coefficient of restitution between any two spheres is \(e\). The first moves with velocity \(u\) towards the second whilst the second and third are at rest. After the first collision the second sphere hits the third after which the velocity of the second sphere is \(u.\) Find \(m_{1}\) in terms of \(m_{2},m_{3}\) and \(e\). deduce that \[ m_{2}e>m_{3}(1+e+e^{2}). \] Suppose that the relation between \(m_{1},m_{2}\) and \(m_{3}\) is that in the formula you found above, but that now the first sphere initially moves with velocity \(u\) and the other two spheres with velocity \(v\), all in the same direction along the line. If \(u>v>0\) use the first part to find the velocity of the second sphere after two collisions have taken place. (You should not need to make any substantial computations but you should state your argument clearly.)

Two particles \(P_{1}\) and \(P_{2}\), each of mass \(m\), are joined by a light smooth inextensible string of length \(\ell.\) \(P_{1}\) lies on a table top a distance \(d\) from the edge, and \(P_{2}\) hangs over the edge of the table and is suspended a distance \(b\) above the ground. The coefficient of friction between \(P_{1}\) and the table top is \(\mu,\) and \(\mu<1\). The system is released from rest. Show that \(P_{1}\) will fall off the edge of the table if and only if \[ \mu<\frac{b}{2d-b}. \] Suppose that \(\mu>b/(2d-b)\) , so that \(P_{1}\) comes to rest on the table, and that the coefficient of restitution between \(P_{2}\) and the floor is \(e\). Show that, if \(e>1/(2\mu),\) then \(P_{1}\) comes to rest before \(P_{2}\) bounces a second time.

A piledriver consists of a weight of mass \(M\) connected to a lighter counterweight of mass \(m\) by a light inextensible string passing over a smooth light fixed pulley. By considerations of energy or otherwise, show that if the weights are released from rest, and move vertically, then as long as the string remains taut and no collisions occur, the weights experience a constant acceleration of magnitude \[ g\left(\frac{M-m}{M+m}\right). \] Initially the weight is held vertically above the pile, and is released from rest. During the subsequent motion both weights move vertically and the only collisions are between the weight and the pile. Treating the pile as fixed and the collisions as completely inelastic, show that, if just before a collision the counterweight is moving with speed \(v\), then just before the next collision it will be moving with speed \(mv/\left(M+m\right)\). {[}You may assume that when the string becomes taut, the momentum lost by one weight equals that gained by the other.{]} Further show that the times between successive collisions with the pile form a geometric progression. Show that the total time before the weight finally comes to rest is three times the time from the start to the first impact.

The identical uniform smooth spherical marbles \(A_{1},A_{2},\ldots,A_{n},\) where \(n\geqslant3,\) each of mass \(m,\) lie in that order in a smooth straight trough, with each marble touching the next. The marble \(A_{n+1},\) which is similar to \(A_{n}\) but has mass \(\lambda m,\) is placed in the trough so that it touches \(A_{n}.\) Another marble \(A_{0},\) identical to \(A_{n},\) slides along the trough with speed \(u\) and hits \(A_{1}.\) It is given that kinetic energy is conserved throughout.

1. Show that if \(\lambda<1,\) there is a possible subsequent motion in which only \(A_{n}\) and \(A_{n+1}\) move (and \(A_{0}\) is reduced to rest), but that if \(\lambda>1,\) such a motion is not possible.
2. If \(\lambda>1,\) show that a subsequent motion in which only \(A_{n-1},A_{n}\) and \(A_{n+1}\) move is not possible.
3. If \(\lambda>1,\) find a possible subsequent motion in which only two marbles move.

A uniform smooth wedge of mass \(m\) has congruent triangular end faces \(A_{1}B_{1}C_{1}\) and \(A_{2}B_{2}C_{2},\) and \(A_{1}A_{2},B_{1}B_{2}\) and \(C_{1}C_{2}\) are perpendicular to these faces. The points \(A,B\) and \(C\) are the midpoints of \(A_{1}A_{2},B_{1}B_{2}\) and \(C_{1}C_{2}\) respectively. The sides of the triangle \(ABC\) have lengths \(AB=AC=5a\) and \(BC=6a.\) The wedge is placed with \(BC\) on a smooth horizontal table, a particle of mass \(2m\) is placed at \(A\) on \(AC,\) and the system is released from rest. The particle slides down \(AC,\) strikes the table, bounces perfectly elastically and lands again on the table at \(D\). At this time the point \(C\) of the wedge has reached the point \(E\). Show that \(DE=\frac{192}{19}a.\)

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Solution: Conservation of energy, tells us that \(2m \cdot g \cdot 4a = 8amg\) is equal to \(\frac12 m v_{wedge}^2 + \frac12(2m)v_{particle}^2\). Conservation of momentum (horizontally) tells us that \(m v_{wedge}+2mv_{particle, \rightarrow} = 0 \Rightarrow v_{particle, \rightarrow} = -\frac12 v_{wedge}\).

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We know that the particle must remain on the slope, and so \(v_{particle,\downarrow} = \frac{4}{3} \frac{3}{2} v_{wedge} = 2v_{wedge}\). In conclusion, we have: \begin{align*} && 8amg &= \frac12 m v_{wedge}^2 + \frac12 (2m)\left ((-\tfrac12 v_{wedge})^2 + (2v_{wedge})^2 \right ) \\ &&&= \frac{19}{4}mv_{wedge}^2 \\ \Rightarrow && v_{wedge}^2 &= \frac{32}{19}ag \end{align*}. To calculate the time the ball bounces for, note that: \(s = ut + \frac12 at^2 \Rightarrow 0 = 2v_{wedge} - \frac12 gt \Rightarrow t = \frac{4v_{wedge}}{g}\). During this time, the wedge (and ball) who horizontally are moving apart with speed \(\frac32 v_{wedge}\) we have they move apart by: \begin{align*} && DE &= \underbrace{\frac32 v_{wedge}}_{\text{speed they move apart}} \cdot \underbrace{\frac{4v_{wedge}}{g}}_{\text{time they are moving apart for}} \\ &&&= \frac{6}{g} v_{wedge}^2 \\ &&&= \frac{6}{g}\frac{32}{19}ag \\ &&&= \frac{192}{19}a \end{align*}

A smooth billiard ball moving on a smooth horizontal table strikes another identical ball which is at rest. The coefficient of restitution between the balls is \(e(<1)\). Show that after the collision the angle between the velocities of the balls is less than \(\frac{1}{2}\pi.\) Show also that the maximum angle of deflection of the first ball is \[ \sin^{-1}\left(\frac{1+e}{3-e}\right). \]

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Solution:

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Set up the coordinate frame so that the \(x\)-direction is the line of centres of the spheres. Then if the initial velocities are \(\displaystyle \binom{u_x}{u_y}\) and \(\displaystyle \binom{0}{0}\). Then the final velocities must be: \(\displaystyle \binom{v_{x1}}{u_y}\) and \(\displaystyle \binom{v_{x2}}{0}\) where \(mu_x = mv_{x1}+mv_{x2}\) by conservation of energy and \(\frac{v_{x1}-v_{x2}}{u_x} = -e\). \begin{align*} && \begin{cases} v_{x1}+v_{x2} &= u_x \\ v_{x1}-v_{x2} &= -eu_x \\ \end{cases} \\ \Rightarrow && 2v_{x1} &= (1-e)u_x \\ \Rightarrow && v_{x1} &= \frac{(1-e)}{2} u_x \\ && v_{x2} &= \frac{1+e}{2} u_x \end{align*} Notice that since \(0 < e < 1\) we must have \(v_{x1} > 0\) and so the ball on the left is still continuing in the positive direction, therefore the angle will be less than \(\frac12 \pi\). The angle the first ball is deflected through is the angle between: \(\displaystyle \binom{u_x}{u_y}\) and \(\displaystyle \binom{\frac{1-e}{2}u_x}{u_y}\). We can scale the velocities so \(u_y = 1\). So we are interested in the angle between \(\displaystyle \binom{x}{1}\) and \(\displaystyle \binom{\frac{1-e}{2}x}{1}\). To maximise \(\theta\) we can maximise \(\tan \theta\), so: \begin{align*} && \tan \theta &= \frac{\frac{2}{(1-e)x-\frac{1}{x}}}{1+\frac{2}{(1-e)x^2}} \\ &&&= \frac{2x-(1-e)x}{(1-e)x^2+2} \\ &&&= \frac{(1+e)x}{(1-e)x^2+2} \\ \\ \frac{\d}{\d t}: &&&= \frac{(1+e)((1-e)x^2+2)-2(1+e)(1-e)x^2}{\sim} \\ &&&= \frac{2(1+e)-(1+e)(1-e)x^2}{\sim}\\ \frac{\d}{\d t} = 0: &&0 &= 2(1+e)-(1+e)(1-e)x^2 \\ \Rightarrow && x &= \pm \sqrt{\frac{2}{1-e}} \\ \\ \Rightarrow && \tan \theta &= \frac{\pm(1+e)\sqrt{\frac{2}{1-e}}}{2+2} \\ &&&= \pm \frac{\sqrt{2}(1+e)}{4\sqrt{1-e}} \\ \Rightarrow && \cot^2 \theta &= \frac{8(1-e)}{(1+e)^2} \\ \Rightarrow && \cosec^2 \theta &= \frac{8(1-e)}{(1+e)^2} + 1 \\ &&&= \frac{8-8e+1+2e+e^2}{(1+e)^2} \\ &&&= \frac{9-6e+e^2}{(1+e)^2} \\ &&&= \frac{(3-e)^2}{(1+e)^2} \\ \Rightarrow && \theta &= \sin^{-1} \left ( \frac{1+e}{3-e}\right) \end{align*}

A smooth sphere of radius \(r\) stands fixed on a horizontal floor. A particle of mass \(m\) is displaced gently from equilibrium on top of the sphere. Find the angle its velocity makes with the horizontal when it loses contact with the sphere during the subsequent motion. By energy considerations, or otherwise, find the vertical component of the momentum of the particle as it strikes the floor.

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Solution:

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Whilst the particle is on the surface of the sphere consider the energy. Letting the height of centre of the sphere by our \(0\) GPE level, the initial energy is \(mgr\) (assuming that the initial speed is so close to \(0\) as to make no difference). When it makes an angle \(\theta\) with the horizontal it's energy will be \(mgr \sin \theta + \frac12 m v^2\). By conservation of energy: \(mgr \sin \theta + \frac12 m v^2 = mgr \Rightarrow v^2 = 2gr(1-\sin \theta)\) \begin{align*} \text{N2}(\text{radially}): && mg \sin \theta - R &= m \frac{v^2}{r} \\ \Rightarrow && R &= mg\sin \theta - \frac{m}{r} 2gr(1-\sin \theta) \\ &&&=mg \l 3\sin \theta - 2 \r \end{align*} Since \(R\) must be positive whilst the particle is in contact with the sphere, the angle \(\theta\) makes with the horizontal when it leaves the sphere is \(\sin^{-1} \frac{2}{3}\). At this point \(v^2 = 2gr(1-\sin \theta) = \frac{2}{3}gr\) Again, considering energy, the initial energy is \(mgr\). The final energy is \(-mgr + \frac12mu_x^2 + \frac12mu_y^2\) When the particle leaves the surface it has speed \(v= \frac23 gr\), so the component \(u_x = \sqrt{v}\sin \theta\). By conservation of energy therefore: \begin{align*} && mgr &= -mgr + \frac12mu_x^2 + \frac12mu_y^2 \\ \Rightarrow && \frac12 u_y^2 &= 2gr - \frac12 u_x^2 \\ &&&= 2gr - \frac12 (\sqrt{v} \sin \theta)^2 \\ &&&= 2gr - \frac12 \frac23gr \sin^2 \theta \\ &&&= 2gr - \frac13gr \frac{4}{9} \\ &&&= \frac{50}{27}gr \\ \Rightarrow && u_y &= \frac{10}{3\sqrt{3}}\sqrt{gr} \end{align*} Therefore vertical component of momentum is \(\displaystyle \frac{10}{3\sqrt{3}}\sqrt{gr}m\)