Problems

Filters
Clear Filters

2 problems found

2010 Paper 2 Q10
D: 1600.0 B: 1516.0
momentum elastic collision conservation of energy conservation of momentum inequality optimisation logarithms chain of collisions

  1. In an experiment, a particle \(A\) of mass \(m\) is at rest on a smooth horizontal table. A particle \(B\) of mass \(bm\), where \(b >1\), is projected along the table directly towards \(A\) with speed \(u\). The collision is perfectly elastic. Find an expression for the speed of \(A\) after the collision in terms of \(b\) and \(u\), and show that, irrespective of the relative masses of the particles, \(A\) cannot be made to move at twice the initial speed of \(B\).
  2. In a second experiment, a particle \(B_1\) is projected along the table directly towards \(A\) with speed \(u\). This time, particles \(B_2\), \(B_3\), \(\ldots\,\), \(B_n\) are at rest in order on the line between \(B_1\) and \(A\). The mass of \(B_i\) (\(i=1\), \(2\), \(\ldots\,\), \(n\)) is \(\lambda^{n+1-i}m\), where \(\lambda>1\). All collisions are perfectly elastic. Show that, by choosing \(n\) sufficiently large, there is no upper limit on the speed at which \(A\) can be made to move. In the case \(\lambda=4\), determine the least value of \(n\) for which \(A\) moves at more than \(20u\). You may use the approximation \(\log_{10}2 \approx 0.30103\).

Solution:

  1. TikZ diagram
    Since the collision is perfectly elastic, the speed of approach and separation are equal, ie \(v_B = v_A - u\) \begin{align*} \text{COM}: && bmu &= bm(v_A - u) + mv_A \\ \Rightarrow && (b+1)v_A &= 2bu \\ \Rightarrow && v_A &= \frac{2b}{b+1} u \end{align*} Since \(0 < \frac{b}{b+1} < 1\), the largest \(0 < v_A < 2u\)
  2. After the first collision with each \(B_i\) we will have \(\displaystyle v_{i+1} = \frac{2\lambda}{\lambda + 1}v_i\), ie \(\displaystyle v_{i+1} = \left (\frac{2\lambda}{\lambda + 1} \right)^i u\) and so \(\displaystyle v_A = \left (\frac{2\lambda}{\lambda + 1} \right)^n u\) which can be arbitrarily large. Suppose \(\lambda = 4\), then \begin{align*} && 20u &< v_A \\ &&&= \left (\frac{8}{5} \right)^n u \\ \Rightarrow && \log_{10} 20 < n \log_{10}(16/10) \\ && \log_{10} 2 + 1 < n 4\log_{10} 2 - n \\ \Rightarrow && n &> \frac{ \log_{10} 2 + 1}{ 4\log_{10} 2 - 1} \\ &&&\approx \frac{0.30103+1}{4 \times 0.30103 -1}\\ &&&= \frac{1.30103}{0.20412} \\ &&&>6 \end{align*} So \(n =7\) is the smallest possible

View
1990 Paper 2 Q14
D: 1600.0 B: 1484.0
momentum collisions elastic collision Newton's experimental law chain of collisions conservation of energy conservation of momentum inequality

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.

Solution: Without loss of generality, let \(m = u = 1\).

  1. \begin{align*} \text{COM}: && 1&= v_n + \lambda v_{n+1} \\ && &= v_n + \lambda v_{n+1}\\ \text{COE}: && \frac12 &= \frac12 v_n^2 + \frac12 \lambda v_{n+1}^2 \\ && 1 &= v_n^2 +\lambda v_{n+1}^2 \\ \\ \Rightarrow && v_n^2 + 2\lambda v_n v_{n+1} + \lambda^2 v_{n+1}^2 &= v_n^2 + \lambda v_{n+1}^2 \\ && \lambda v_{n+1}^2 &= v_{n+1}^2 - 2 v_n v_{n+1} \\ && \lambda v_{n+1} &= (v_{n+1} - 2v_n) \\ && (1-\lambda)v_{n+1} &= 2v_n \end{align*} Since \(v_{n+1} > v_n > 0\) this is only possible if \(\lambda < 1\)
  2. \begin{align*} \text{COM}: && 1&= v_{n-1}+v_n+\lambda v_{n+1} \\ && 1&= v_{n-1} + v_n + \lambda v_{n+1} \\ \text{COE}: && \frac12 &= \frac12 v_{n-1}^2+\frac12v_n^2+\frac12\lambda v_{n+1}^2 \\ && 1&= v_{n-1}^2 + v_n^2 + \lambda v_{n+1}^2 \\ \\ \Rightarrow && 1 &= v_{n-1}^2 + v_n^2 + \lambda \frac{(1-v_{n-1}-v_n)^2}{\lambda^2} \\ &&&= v_{n-1}^2 + v_n^2 + \frac{(1-v_{n-1}-v_n)^2}{\lambda} \\ \Rightarrow && 1 &< v_{n-1}^2 + v_n^2 + (1-v_{n-1}-v_n)^2 \\ &&&= 2v_{n-1}^2+2v_n^2 + 1-2v_{n-1}-2v_{n-2} +2v_{n-1}v_n\\ \Rightarrow && v_{n-1}+v_n & <(v_{n-1}+v_n)^2 - v_{n-1}v_n \end{align*} but this cannot be true since \(0 < v_{n-1}+v_n < 1\) and \(v_n v_{n-1} > 0\)
  3. The only way this is possible is if the first and last marble are moving. \begin{align*} \text{COM}: && 1 &= v_0 +\lambda v_{n+1} \\ \text{COE}: && \frac12 &= \frac12 v_0^2 + \frac12 \lambda^2 v_{n+1} \\ && 1 &= v_0^2 + \lambda v_{n+1}^2 \\ \Rightarrow && 2v_0 + \lambda v_{n+1} &= v_{n+1} \\ \Rightarrow && v_{n+1} &=\frac{2}{1-\lambda} v_0 \\ \Rightarrow && v_0 &= \frac{1-\lambda}{1+\lambda} \\ && v_{n+1} &= \frac{2}{1+\lambda} \end{align*} which will work since \(v_0\) can travel backwards.

View