Problems

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Solution:

1. Just before collision \(n-1\): Velocity of \(P\) is \(u_{n-2}\) Velocity of \(Q\) is \(-v_{n-2}\) \begin{align*} COM: && mu_{n-2}+km(-v_{n-2}) &= mu_{n-1}+kmv_{n-1} \\ \Rightarrow && u_{n-2}-kv_{n-2} &= u_{n-1}+kv_{n-1} \\ NEL: && v_{n-1}-u_{n-1} &= -e((-v_{n-2})-u_{n-2}) \\ \Rightarrow && v_{n-1}-u_{n-1} &= e(v_{n-2}+u_{n-2}) \end{align*} \begin{align*} &&kv_{n-1} &= u_{n-2} - kv_{n-2}-u_{n-1} \\ &&kv_{n-1}&= ku_{n-1}+kev_{n-2}+keu_{n-2} \\ \Rightarrow && kv_{n-2}(1+e) &= u_{n-2}(1-ke)-u_{n-1}(1+k) \\ \Rightarrow && kv_{n-1}(1+e) &= u_{n-1}(1-ke)-u_{n}(1+k) \\ && k(1+e)v_{n-1}-k(1+e)u_{n-1} &= k(1+e)e(v_{n-2}+u_{n-2}) \\ \Rightarrow && u_{n-1}(1-ke)-u_{n}(1+k)-k(1+e)u_{n-1} &= e(u_{n-2}(1-ke)-u_{n-1}(1+k))+k(1+e)eu_{n-2} \\ \Rightarrow && 0 &= (1+k)u_n + ((ke-1)+k(1+e)-e(1+k))u_{n-1} \\ &&& \quad \quad + (e(1-ke)+k(1+e)e)u_{n-2} \\ \Rightarrow && 0 &= (1+k)u_n- (1-k)(1+e)u_{n-1} +e(1+k)u_{n-2} \end{align*}
2. \(u_0 = A + B\) \begin{align*} &&& \begin{cases}u_0 - kv_0 &= kv_1 + u_1 \\ \frac12 (u_0+v_0) &= v_1 - u_1 \\ \end{cases} \\ \Rightarrow && (1+k)u_1 &= u_0 - kv_0 - \frac{k}{2}(u_0 + v_0) \\ \Rightarrow && u_1 &= \frac{1}{k+1} \l u_0 (1-\frac{k}{2}) - \frac32 k v_0 \r \\ &&&= \frac{67}{70} u_0 - \frac{3}{70} v_0 \end{align*} Therefore \(A+B = u_0, \frac{49A+50B}{70} = \frac{67}{70} u_0 - \frac{3}{70} v_0\) \begin{align*} && A+B &= u_0 \\ && 49A+50B &= 67u_0 - 3v_0 \\ \Rightarrow && 50u_0 - A &= 67u_0 - 3v_0 \\ \Rightarrow && A &= -17u_0 + 3v_0 \\ && B &= 18u_0 - 3v_0 \end{align*} If \(0 < 6u_0 < v_0\), then \(B < 0\) and as \(n \to \infty\) we will find that \(\l \frac57 \r^n\) dominates \(\l \frac7{10} \r^n\) and so our velocity will be negative and the particle will change direction

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Solution:

1. 

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   \begin{align*} \text{COM}: && mu &= mv + \lambda m u_1 \\ \Rightarrow && u &= v + \lambda u_1 \tag{1} \\ \text{NEL}: && e &= \frac{u_1-v}{u} \\ \Rightarrow && eu &= u_1 - v \tag{2} \\ (1)+(2) && (1+e)u &= (1+\lambda) u_1 \\ \Rightarrow && u_1 &= \frac{1+e}{1+\lambda}u \\ && v &= u_1 - eu \\ &&&= \frac{1+e - (1+\lambda)e}{1+\lambda} u \\ &&&= \frac{1-\lambda e}{1+\lambda}u \end{align*} Note that subsequent (first (and second)) are the same as these, therefore: \begin{align*} u_n &= \left ( \frac{1+e}{1+\lambda} \right)^n u \\ v_n &= \frac{1-\lambda e}{1+\lambda } u_n \\ &= \frac{1-\lambda e}{1+\lambda } \left ( \frac{1+e}{1+\lambda} \right)^n u \end{align*}
2. If \(e > \lambda\) then \((1-\lambda e) > 1-e^2 > 0\) and \begin{align*} \frac{v_{n+1}}{v_n} &= \frac{1+e}{1+\lambda} > 1 \end{align*} So the particles are moving away from each other - hence no more collisions.
3. If \(e = \lambda\) then \(u_n = u\) and \(v_n = (1-\lambda)u\) so all the particles end up moving at the same speed. \begin{align*} \text{initial k.e.} &= \frac12 m u^2 \\ \text{final k.e.} &= \frac12 m((1-e)u)^2 + \sum_{n = 1}^{\infty} \frac12 \lambda^n m ((1-e)u)^2 \\ &= \frac12mu^2(1-e)^2 \left ( \sum_{n=0}^{\infty} e^n \right) \tag{\(e = \lambda\)} \\ &= \frac12 mu^2(1-e)^2 \frac{1}{1-e} \\ &= \frac12m u^2 (1-e) \\ \text{change in k.e.} &= \frac12 m u^2 - \frac12m u^2 (1-e) \\ &= e\frac12m u^2 \end{align*} Ie the total energy lost approaches a fraction of \(e\).
4. If \(\lambda e = 1\), after the second collision the particle will be stationary. ie \begin{align*} \text{initial k.e.} &= \frac12 m u^2 \\ \text{k.e. after }n\text{ collisions} &= \frac12 \lambda^n m \left (\left ( \frac{1+e}{1+\lambda} \right)^n u \right)^2\\ &= \frac12 \lambda^n m \left ( \frac{1+\frac1{\lambda}}{1+\lambda} \right)^{2n} u&2\\ &= \frac12 \lambda^n m \left ( \frac{1+\frac1{\lambda}}{1+\lambda} \right)^{2n} u\\ &= \frac12 \lambda^n m \left ( \frac{1}{\lambda} \right)^{2n} u\\ &= \frac12 m \lambda^{-n} u\\ &\to 0 \end{align*} Eventually we lose all the kinetic energy.

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Solution:

Collision between A & B. Since the speed of approach is \(u\) and the coefficient of restitution is \(e\) we must have \(v_B = v_A + eu\). \begin{align*} \text{COM}: && \lambda m u &= \lambda m (v_B - eu) + m v_B \\ \Rightarrow && v_B(\lambda + 1) &=\lambda (1+ e) u \\ \Rightarrow && v_B &= \frac{\lambda(1+ e)}{1+\lambda} u \end{align*} Collision between A & B. Since the speed of approach is \(u\) and the coefficient of restitution is \(e\) we must have \(v_D = v_C + eu\). \begin{align*} \text{COM}: && m(-u) &= mv_C + m(v_C + eu) \\ \Rightarrow && 2v_C &= -(1+e)u \\ \Rightarrow && v_C &= -\frac{1+e}{2} u \end{align*}

1. 

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   \begin{align*} \text{NEL}: && w_C &= e(v_B - v_C) \\ \text{COM}: && mv_B+ mv_C &= m w_C \\ \Rightarrow && w_C &= v_B + v_C\\ \Rightarrow && e(v_B - v_C) &= (v_B + v_C) \\ \Rightarrow && (1-e)v_B &= -(1+e)v_C \\ \Rightarrow && (1-e) \frac{\lambda(1+ e)}{1+\lambda} &= (1+e) \frac{1+e}{2} \\ \Rightarrow && 2\lambda - 2\lambda e &= 1+\lambda + e + \lambda e \\ \Rightarrow && (3\lambda +1)e &= \lambda - 1 \\ \Rightarrow && e &= \frac{\lambda -1}{3\lambda + 1} \\ &&&< \frac{\lambda - 1 + \frac{4}{3}}{3\lambda + 1} \\ &&& = \frac13 \end{align*}
2. Since they move towards each other at the same speed \(w_C = - v_D\) \begin{align*} && w_C &= - v_D \\ \Rightarrow && v_B + v_C &= -(v_C+eu) \\ \Rightarrow && -eu &= v_B +2v_C \\ &&&= \frac{\lambda(1+ e)}{1+\lambda} u -(1+e)u \\ \Rightarrow && 1 &= \frac{\lambda(1+e)}{1+\lambda} \\ \Rightarrow && 1+\lambda &= \lambda \left ( 1 + \frac{\lambda -1}{3\lambda+1} \right) \\ &&&= \lambda \frac{4\lambda}{3\lambda +1} \\ \Rightarrow && 1+4\lambda + 3\lambda^2 &= 4\lambda^2 \\ \Rightarrow && 0 &= \lambda^2 - 4\lambda - 1 \\ \Rightarrow && \lambda &= \frac{4 \pm \sqrt{20}}{2} \\ &&&= 2\pm \sqrt{5} \\ \Rightarrow && \lambda &= 2 + \sqrt{5} \\ && e &= \frac{1+\sqrt{5}}{7+3\sqrt{5}} \\ &&&=\sqrt{5}-2 \end{align*}

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Solution:

1. Since \(R\) is constant, \(F=ma \Rightarrow \text{acc} = \frac{R}{m}\) and \(v^2 = u^2 + 2as\) so \(0 = u^2 - 2 \frac{R}{m}a\), ie \(a = \frac{m u^2}{2R}\)
2. By conservation of momentum, the bullet/block combination will eventually be travelling at \(v = \frac{m}{m+M}u\). The bullet will slow down to this speed in a time of \(\frac{m}{m+M}u = u - \frac{R}{m} T \Rightarrow T = \frac{Mm}{R(m+M)}u\) and will travel \(b+c = \frac{\left ( 1 - \left ( \frac{m}{m+M} \right)^2\right)u^2m}{2R}= \left ( 1 - \left ( \frac{m}{m+M} \right)^2\right)a\). The block will travell \(c = \frac12 \frac{R}{M} \frac{M^2m^2u^2}{R^2(m+M)^2} = \frac{Mm}{(m+M)^2}a\) Therefore the \(b = \left ( 1 - \left ( \frac{m}{m+M} \right)^2\right)a - \frac{Mm}{(m+M)^2}a = \frac{M}{M+m}a\)

Work done by friction is the relative gain in KE for the block. ie \(R \cdot c = \frac12 M \left ( \frac{m}{m+M}u\right)^2 \Rightarrow c = \frac{Mm}{(M+m)^2}a\).

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Solution:

1. The story for first ball (before it collides with the second ball) will be exactly the same, ie it rebounds with speed \(eV\). For the second ball, it will also have fallen a distance \(H\) and will be travelling with the same speed \(V\). Their speed of approach therefore will be \((1+e)V\), and the speed of separating therefore must be \(e(1+e)V\) Given the centre of the second ball reaches a height of \(h\) (from a position of height) \(2R+r\), we must have: \begin{align*} && v^2 &= u^2 + 2as \\ && 0 &= w^2 - 2g(h - 2R-r) \\ \Rightarrow && w^2 &= 2g(h-2R-r) \end{align*} Taking upwards to be positive, then we have:
   

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   \begin{align*} \text{COM}: && MeV + m(-V) &= M(w-e(1+e)V) + mw \\ \Rightarrow && V(Me-m)+e(1+e)MV &= w(M+m) \\ \Rightarrow && w &= \frac{2Me+e^2M-m}{M+m} V \\ \Rightarrow && w^2 &= \left ( \frac{M/m(2e+e^2)-1}{M/m+1}\right)^2 V^2 \\ \Rightarrow && 2g(h-2R-r) &= \left ( \frac{M/m(2e+e^2)-1}{M/m+1}\right)^2 2gH \\ \Rightarrow && \left ( \frac{M/m(2e+e^2)-1}{M/m+1}\right)^2 &= \frac{h-2R-r}{H} \\ &&&= \frac{4.5-0.4-0.05}{1.8} \\ &&&= \frac{9}{4} \\ \Rightarrow && \frac{M/m(\frac43+\frac49)-1}{M/m+1} &= \frac32 \\ \Rightarrow && \frac{16}9M/m-1 &= \frac32 M/m+\frac32 \\ \Rightarrow && \frac{5}{18}M/m &= \frac{5}{2} \\ \Rightarrow && M/m &= 9 \end{align*}

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Solution:

1. First Collision:
   

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   By NEL, \(v_2 = v_1 + eu\), so \begin{align*} \text{COM}: && mu &= mv_1 + m(v_1 + eu) \\ \Rightarrow && 2mv_1 &= mu(1-e) \\ \Rightarrow && v_1 &= \frac12 u(1-e) \\ && v_2 &= \frac12 u(1-e) + eu \\ &&&= \frac12 u(1+e) \end{align*} The second collision is identical to the first except replacing \(u\) with \(\frac12u(1+e)\), therefore after that collision: \begin{align*} && \text{first particle} &= \frac12 u(1-e) \\ && \text{second particle} &= \frac12 \left (\frac12 u(1+e) \right)(1-e) \\ &&&= \frac14 u(1-e^2) \\ && \text{third particle} &= \frac12 \left (\frac12 u(1+e) \right)(1+e) \\ &&&= \frac14 u(1+e)^2 \end{align*} After all these collisions, all particles are moving in the same direction (since they all have positive velocity), but the first particle is now travelling faster than the second particle (as \(\frac12(1-e) < 1\)). Therefore they will collide again.
2. The third collision:
   

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   The speed of approach will be \(\frac12u(1-e) - \frac14u(1-e^2) = \frac14u(1-e)(2 - (1+e)) = \frac14 u(1-e)^2\), therefore by NEL, \(w_2 = w_1 + \frac14ue(1-e)^2\) \begin{align*} \text{COM}: && m\frac12u(1-e) + m \frac14u(1-e^2) &= mw_1 + m\left (w_1 + \frac14ue(1-e)^2 \right) \\ \Rightarrow && \frac14u(1-e)(2+(1+e)) &= 2w_1 + \frac14ue(1-e)^2 \\ \Rightarrow && 2w_1 &= \frac14u(1-e)(3+e)-\frac14ue(1-e)^2 \\ &&&= \frac14u(1-e)(3+e-e(1-e)) \\ &&&= \frac14u(1-e)(3+e^2) \\ \Rightarrow && w_1 &= \frac18 u(1-e)(3+e^2) \\ && w_2 &= \frac18 u(1-e)(3+e^2) + \frac14ue(1-e)^2 \\ &&&= \frac18u(1-e)(3+e^2+2e(1-e)) \\ &&&= \frac18u(1-e)(3+2e-e^2) \\ &&&= \frac18u(1-e)(1+e)(3-e) \\ \end{align*} A fourth collision is possible, iff \begin{align*} && \frac18u(1-e)(1+e)(3-e)&> \frac14 u(1+e)^2 \\ \Leftrightarrow && (1-e)(3-e)&> 2 (1+e) \\ \Leftrightarrow &&3-4e-e^2&> 2+2e \\ \Leftrightarrow &&1-5e-e^2&>0 \\ \Leftrightarrow && e &< 3-\sqrt{2} \end{align*}

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Solution: \begin{align*} \text{COM}: && \sum_{i=1}^n im \cdot \frac{u}{i} &= \left ( km + \sum_{i=1}^n im \right) v \\ \Rightarrow && nu &= \left ( k + \frac{n(n+1)}{2} \right) v \\ \Rightarrow && v &= \frac{2nu}{2k + n(n+1)} \end{align*} If \(2k = N(N+1)\), there will be no more collisions when \(v_n > \frac{u}{n+1}\), ie \begin{align*} && \frac{u}{n+1} &<\frac{2nu}{2k + n(n+1)} \\ \Leftrightarrow && N(N+1) + n(n+1) &< 2n(n+1) \\ \Leftrightarrow && N(N+1) &< n(n+1) \\ \end{align*} Therefore \(n = N+1\) and there will be \(N+1\) collisions. The loss of kinetic energy is: \begin{align*} && \text{initial k.e.} &= \sum_{k=1}^{N+1} \frac12 im \cdot \frac{u^2}{i^2} \\ &&&= \frac12 m u^2 \left ( \sum_{k=1}^{N+1} \frac{1}{i}\right) \\ && \text{final k.e.} &= \frac12 \left ( k + \frac{(N+1)(N+2)}{2}\right)m \left ( \frac{2(N+1)u}{N(N+1)+(N+1)(N+2)} \right)^2 \\ &&&= \frac12 m u^2 \frac{2(N+1)^2}{(N+1)(2N+2)} \\ &&&= \frac12 mu^2 \\ \Rightarrow && \Delta \text{ k.e.} &= \frac12 m u^2 \left ( \sum_{k=2}^{N+1} \frac{1}{i}\right) \end{align*}

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Solution:

Since the coefficient of restitution is \(e\) and the speed of approach is \(3u\), \(v_B = v_A + 3eu\), \begin{align*} \text{COM}: && 2m\cdot2u + m \cdot (-u) &= 2m v_A + m(v_A + 3eu) \\ \Rightarrow && 3u &= 3v_A + 3eu \\ \Rightarrow && v_A &= (1-e)u \\ \Rightarrow && v_B &= (1+2e)u \end{align*} After rebounding from the wall, the velocity of \(B\) will be \(-fv_B\). So for the second collision (between the particles) we will have: \begin{align*} \text{NEL}: && w_B - w_A &= e((1-e)u+(1+2e)fu) \\ \Rightarrow && w_B - w_A &= (1-e+f+2ef)eu \tag{1} \\ \text{COM}: && 2m w_A + w_B &= 2m(1-e)u -m(1+2e)fu \\ \Rightarrow && 2w_A + w_B &= (2-2e -f-2ef)u \tag{2} \\ (2) + 2\times(1): && 3w_B &= (2-2e -f-2ef)u +2(1-e+f+2ef)eu \\ &&&= (2-2e-f-2ef)u+(2e-2e^2+2ef+4e^2f)u \\ &&&= (2-2e^2-f+4e^2f)u \\ &&&= 2(1-e^2)-f(1-4e^2)u \\ \Rightarrow && w_B &= \frac23 (1-e^2)u-\frac13(1-4e^2)fu \end{align*} Since we've always taken towards the wall as positive, the question is whether or not this is positive for all values of \(e\) and \(f\). The first term is clearly positive, so in order to have a chance of being negative, we must have that \(1-4e^2 > 0\) and \(f\) is as large as possible, so wlog \(f = 1\). \begin{align*} 2-2e^2-1+4e^2 = 1+2e^2 > 0 \end{align*} \end{align*}

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Solution:

1. \begin{align*} \text{COM}: && mu + Mv &= mp + Mq \\ \Rightarrow && m(u-p) &= M(q-v) \\ \text{NEL}: && q-p &= e(u-v) \\ && q +ev &= p+eu \\ && \Delta \text{ k.e.} &= \frac12 m u^2 + \frac12 M v^2 -\frac12 m p^2 - \frac12 M q^2 \\ &&&= \frac12m (u^2 - p^2)+\frac12M(v^2-q^2) \\ &&&= \frac12m (u^2 - p^2)+\frac12M(v-q)(v+q) \\ &&&= \frac12m(u^2-p^2) - \frac12 m(u-p)(v+q) \\ &&&= \frac12 m(u-p) \left ( u+p-v-q\right) \\ &&&= \frac12 m(u-p) \left (u-v+(p-q)\right) \\ &&&= \frac12 m(u-p) \left (u-v-e(u-v)\right) \\ &&&= \frac12m(u-p)(u-v)(1-e) \end{align*} Since the loss in energy is positive, and \(m\), \(u-v\) and \(1-e\) are all positive, so is \(u-p\), ie \(u \geq p\)
2. \begin{align*} && \frac12 m u^2 - \frac12mp^2 &= \frac12Mv^2 - \frac12Mq^2 \\ && \frac12 m(u-p)(u+p) &= \frac12 M (v-q)(v+q) \\ && \frac12 m (u-p)(u+p) &= -\frac12 m(u-p)(v+q) \\ \Rightarrow && u+p+v+q &= 0 \end{align*} \begin{align*} && p+q &= -(u+v)\\ &&mp+Mq &= mu+Mv \\ \Rightarrow && (M-m)q &= mu+Mv+mu+mv\\ \Rightarrow && q &= \frac{(M+m)v+2mu}{M-m} \\ \Rightarrow && (m-M)p &= mu+Mv+Mu+Mv \\ \Rightarrow && p &= -\frac{(M+m)u+2Mv}{M-m} \\ \\ && e &= \frac{q-p}{u-v} \\ &&&= \frac{(M+m)v+2mu+(M+m)u+2Mv}{(u-v)(M-m)} \\ &&&= \frac{(3M+m)v+(3m+M)u}{(u-v)(M-m)} \end{align*}

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Solution:

1. After the first collision, it takes \(B\), \(\frac{d}{v_B} = \frac{d}{u} \frac{2(k+1)}{3} = \frac{4d}{3u}\) to collide with \(C\). During which time \(B\) and \(A\) have been moving apart with speed \(\frac12u\) and so are a distance \(\frac{2d}{3}\) apart. After the second collision, \(w_B = \frac{3(4\cdot 1 - 3)}{8(1+1)(1+3)}u = \frac{3}{64}u\) and \(v_A = \frac{1}{4}u\) so they are moving together at speed \(\frac{16-3}{64}u = \frac{13}{64}u\). It will take them \(\frac{2d}{3} \div \frac{13}{64}u = \frac{128d}{3 \times 13u}\) to do this for a total time of \(\frac{128d}{3 \times 13u} + \frac{4d}{3u} = \frac{(128+52)d}{3 \times 13 u} = \frac{60d}{13u}\)