Problems

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Solution: Since the particles are identical and are projected with the same speed, the only way they can reach the same point \(x\) at the same time, is if \(A\) has reached it's apex and started descending. Considering \(P\), we must have (setting the level of \(A\) to be the \(0\) G.P.E. level), suppose it travels a distance \(x\) before becoming stationary: \begin{align*} \text{N2}(\nwarrow): && R - 4g \cos(30^\circ) &= 0 \\ \Rightarrow && R &= 20\sqrt{3} \\ \Rightarrow && \mu R &= \frac1{5 \sqrt{3}} (20 \sqrt{3}) \\ &&&= 4 \\ \end{align*} Therefore in the two phases of the journey the particle is being accelerated down the slope by either \(6\) or \(4\). \(v^2 = u^2 + 2as \Rightarrow 0 = 36 - 12s \Rightarrow s = 3\). \(v = u + at \Rightarrow t = 1\). Therefore after \(1\) second \(P\) reaches its highest point having travelled \(3\) metres. It will pass back to the start in \(s = ut + \frac12 a t^2 \Rightarrow 3 = \frac12 4 t^2 \Rightarrow t = \sqrt{3/2}\) seconds, ie the constraint is that the particle hasn't already past \(Q\) before the collision. The collision will occur when \(s = 6t - \frac12 6 t^2\) and \(s =3 - \frac12 4 (t+T-1)^2\) coincide, ie: \begin{align*} && 6t - 3t^2 &= 3 - 2(t+T-1)^2 \\ && 0 &= 3 -2(T-1)^2 -(4(T-1)+6)t + t^2 \\ && 0 &= 3 -2(T-1)^2 -(4T+2)t + t^2 \\ \Rightarrow && t &= \frac{4T+2 \pm \sqrt{(4T+2)^2 - 4(3-2(T-1)^2)}}{2} \\ &&&= \frac{4T+2 \pm \sqrt{24T^2}}{2} \\ &&&= 2T + 1 \pm \sqrt{6} T \\ &&&= (2 \pm \sqrt{6})T + 1 \end{align*} we must take the smaller root, ie \((2-\sqrt{6})T + 1\). In the case the collision occurs exactly at the start, the particle \(P\) has traveled \(6\) meters, against a force of \(4\) newtons of friction, ie work done is \(24\) Joules.

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Solution: \begin{align*} \text{COM}: && \mathbf{v}_1+\mathbf{v}_2 &= \mathbf{v}_1'+\mathbf{v}_2' \tag{1}\\ \text{COE}: && \mathbf{v}_1\cdot\mathbf{v}_1+\mathbf{v}_2\cdot\mathbf{v}_2 &= \mathbf{v}_1'\cdot\mathbf{v}_1'+\mathbf{v}_2'\cdot\mathbf{v}_2' \tag{2} \\ \\ (1): && (\mathbf{v}_1+\mathbf{v}_2 )\cdot(\mathbf{v}_1+\mathbf{v}_2 ) &= (\mathbf{v}_1'+\mathbf{v}_2' )\cdot(\mathbf{v}_1'+\mathbf{v}_2' ) \\ \Rightarrow && \mathbf{v}_1 \cdot \mathbf{v}_2 &= \mathbf{v}_1'\cdot \mathbf{v}_2' \\ && \text{Initial relative speed}^2 &= |\mathbf{v}_1 - \mathbf{v}_2|^2 \\ &&&= (\mathbf{v}_1 - \mathbf{v}_2) \cdot (\mathbf{v}_1 - \mathbf{v}_2) \\ &&&= \mathbf{v}_1\cdot \mathbf{v}_1 - 2 \mathbf{v}_1\cdot \mathbf{v}_2 + \mathbf{v}_2\cdot \mathbf{v}_2 \\ &&&= \mathbf{v}_1'\cdot\mathbf{v}_1'+\mathbf{v}_2'\cdot\mathbf{v}_2' -2 \mathbf{v}_1\cdot\mathbf{v}_2\\ &&&= \mathbf{v}_1'\cdot\mathbf{v}_1'+\mathbf{v}_2'\cdot\mathbf{v}_2' -2 \mathbf{v}_1'\cdot\mathbf{v}_2'\\ &&&= | \mathbf{v}_1'-\mathbf{v}_2'|^2 \\ &&&= \text{Final relative speed}^2 \end{align*} Since \(\mathbf{v}_1 \cdot 0 = 0\) we must have \(\mathbf{v}_1'\cdot\mathbf{v}_2' = \mathbf{v}_1\cdot0 = 0\) therefore their final velocities are perpendicular. We now must have \begin{align*} \text{COM}: && \mathbf{v}_1+\mathbf{v}_2 &= \mathbf{v}_1'+\mathbf{v}_2' \tag{3}\\ \Delta\text{E}: && (1-k)(\mathbf{v}_1\cdot\mathbf{v}_1+\mathbf{v}_2\cdot\mathbf{v}_2) &= \mathbf{v}_1'\cdot\mathbf{v}_1'+\mathbf{v}_2'\cdot\mathbf{v}_2' \tag{4} \\ \\ && 0 + \mathbf{v}_2 &= \mathbf{v}_1' + \mathbf{v}_2' \\ \Rightarrow && V^2 &= ( \mathbf{v}_1' + \mathbf{v}_2' ) \cdot ( \mathbf{v}_1' + \mathbf{v}_2' ) \\ &&&= \mathbf{v}_1'\cdot\mathbf{v}_1'+\mathbf{v}_2'\cdot\mathbf{v}_2' +2 \mathbf{v}_1'\cdot \mathbf{v}_2' \\ &&&= (1-k)V^2 + 2 (\mathbf{v}_2-\mathbf{v}_2') \cdot \mathbf{v}_2' \\ &&&= (1-k)V^2 + 2 \mathbf{v}_2 \cdot \mathbf{v}_2'-2\mathbf{v}_2'\cdot \mathbf{v}_2' \\ &&&= (1-k)V^2 + 2Vx \cos \theta - 2x^2 \\ \Rightarrow && 0 &= -kV^2 + 2Vx \cos \theta -2x^2 \\ \Delta \geq 0: && 0 &\leq 4V^2 \cos^2 \theta -8kV^2 \\ \Rightarrow && k &\leq \frac12\cos^2\theta \leq \frac12 \end{align*}

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Solution: \begin{align*} && \text{energy when projected} &= \frac12 m(2kgh) \\ &&&= kghm \\ && \text{energy when hitting ceiling the first time} &= mgh + \frac12 m v^2 \\ \text{COE}: && kghm &= mgh + \frac12 mv^2 \\ \Rightarrow && v^2 &= 2gh(k-1) \end{align*} It will rebound with speed \(\sqrt{2gh(k-1)}a\). \begin{align*} && \text{energy when rebounding from ceiling} &=gh(k-1)a^2 + mgh \\ && \text{energy before hitting the floor} &= \frac12 mv^2 \\ \text{COE}: && gh(k-1)a^2 + mgh &= \frac12 mv^2 \\ \Rightarrow && v^2 &= 2gh((k-1)a^2+1) \end{align*} The ball will rebound with kinetic energy \(m gh((k-1)a^2+1)a^2 = mgh((k-1)a^4+a^2)\) And will reach the ceiling with kinetic energy \(mgh((k-1)a^4+a^2-1)\). When \(n = 1\), the kinetic energy (before hitting the ceiling for the first time) is \(mgh(k-1)\). Suppose \(s_n\) is the expression for the kinetic energy divided by \(mgh\), ie \(s_1 = k-1\), then: Clearly \(s_1 = k-1 = a^{4\cdot1-4}(k-1) + \frac{a^{4\cdot-4}-1}{a^2+1}\), so our hypothesis holds for \(n=1\). Suppose it is true for \(n\), then the \(n+1\)th time it will be: \begin{align*} s_{n+1} &= s_n a^4+a^2-1 \\ &= \left ( a^{4n-4}(k-1)+\frac{a^{4n-4}-1}{a^{2}+1} \right) a^4 + a^2 - 1 \\ &= a^{4(n+1)-4}(k-1) + \frac{a^{4(n+1)-4}-a^4}{a^2+1} + \frac{a^4-1}{a^2+1} \\ &= a^{4(n+1)-4}(k-1) + \frac{a^{4(n+1)-4}-a^4+a^4-1}{a^2+1} \\ &= a^{4(n+1)-4}(k-1) + \frac{a^{4(n+1)-4}-1}{a^2+1} \end{align*} Which is our desired expression, therefore it is true by induction. We wont reach the ceiling if this energy is not positive, ie: \begin{align*} && 0 &\leq a^{4n-4}(k-1)+\frac{a^{4n-4}-1}{a^{2}+1} \\ \Rightarrow && \frac{1}{a^2+1}&\geq a^{4n-4}\left (k - 1 + \frac{1}{a^2+1} \right) \\ \Rightarrow && a^{4n-4} &\geq \frac{1}{a^2+1} \cdot \frac{1}{k - 1 + \frac{1}{a^2+1}} \\ \Rightarrow && a^{4n-4} &\geq \frac{1}{(k-1)(a^2+1)+1} \\ \Rightarrow && 4(n-1) \ln a &\geq - \ln[(k-1)(a^2+1)+1] \\ \underbrace{\Rightarrow}_{\ln a < 0} && (n-1) &\leq \frac{ - \ln[(k-1)(a^2+1)+1]}{4\ln a} \\ \Rightarrow && n & \leq 1 -\frac{ \ln[(k-1)(a^2+1)+1]}{4\ln a} \\ &&&= 1 -\frac{ \ln[(k-1)a^2+k]}{4\ln a} \end{align*}

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Solution:

Considering the pulley system with the lift (now of mass \(M-m\)) and the counterweight of mass \(M\). Once they start moving, since they are connected by a light inextensible string they must move with the same speed (and by extension the same acceleration). (Up to sign) \begin{align*} \text{N2(lift,}\uparrow):&&(M-m)a &= T-(M-m)g \\ \text{N2(couterweight,}\downarrow):&&Ma &= Mg - T \\ \Rightarrow && (2M-m)a &= mg \\ \Rightarrow && a &= \frac{mg}{2M-m} \end{align*} We could treat the situation as the tile travelling a distance of \(h\) with acceleration \(\displaystyle g \left ( 1 + \frac{m}{2M-m} \right) = g \frac{2M}{2M-m}\). \begin{align*} t &= \sqrt{\frac{2h}{g \frac{2M}{2M-m}}} \\ &= \sqrt{\frac{(2M-m)h}{Mg}} \\ \end{align*} Since the collision between the lift and tile is perfectly inelastic, they immediately coalesce. There is also an impulse in the pulley system, which goes over the pulley, which means there is an impulse acting vertically on the lift and the counterweight. Assume afterwards the lift (and tile) is travelling upwards with speed \(V\) and the counterweight is travelling downwards with speed \(V\) (ie velocity \(-V\)). \begin{align*} \text{for the lift/tile}: && I_{LC} &= (\text{momentum after}) - (\text{momentum before})\\ &&&= MV - ((M-m)at +m(-g)t) \\ &&&= MV - Mat + m(a-g)t \\ \text{for the counterweight}: && I_{LC} &= (\text{momentum after}) - (\text{momentum before})\\ &&&= M(-V) - (M(-a)t) \\ &&&= -MV +Mat \\ \Rightarrow && 2MV &= m(g-a)t + 2Mat \\ &&&= t (2Ma -ma+mg) \\ &&&= 0 \\ \Rightarrow && V &= 0 \end{align*} Therefore, the lift ends up stationary. The energy lost in the collision is: \begin{align*} && E &= KE_{before} - KE_{after} \\ &&&= \underbrace{\frac12 (M-m)a^2 t^2}_{lift} + \underbrace{\frac12 mg^2 t^2}_{tile} + \underbrace{\frac12 Ma^2 t^2}_{counterweight} - \underbrace{0}_{\text{everything is at rest after}} \\ &&&= \frac12 \l (M-m)a^2 + mg^2 + Ma^2 \r t^2 \\ &&&= \frac12 \l 2Ma^2-ma^2 + mg^2 \r t^2 \\ &&&= \frac12 \l (2M-m)a^2 + mg^2 \r t^2 \\ &&&= \frac12 \l mga + mg^2 \r t^2 \\ &&&= \frac12 mg (a + g)t^2 \\ &&&= \frac12 mg \left ( \frac{mg}{2M-m} + g\right ) \frac{(2M-m)h}{Mg} \\ &&&= \frac12 mg \left ( \frac{mg +2Mg - mg}{2M-m} \right) \frac{(2M-m)h}{Mg} \\ &&&= mgh \end{align*} as required.