Day: March 1, 2024

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202403011713 Revision Paper IV Q8

Ball A (mass 3\,\mathrm{kg}) is travelling due east in a straight line with speed 5\,\mathrm{m\,s^{-1}} when it collides with ball B (mass 4\,\mathrm{kg}) which was at rest. After the collision both balls move east with relative speed 2.5\,\mathrm{m\,s^{-1}}. Find their separate speeds.

Extracted from A. Godman & J. F. Talbert. (1973). Additional Mathematics Pure and Applied in SI Units.

Roughwork.

Take eastward positive.

\begin{aligned} \textrm{CoM:}\qquad p_{\textrm{initial}} & = p_{\textrm{final}} \\ \sum_{i=A}^{B}m_iu_i & = \sum_{i=A}^{B}m_iv_i \\ m_Au_A+m_Bu_B & = m_Av_A+m_Bv_B \\ (3)(5)+(4)(0) & = (3)(v_A)+(4)(v_B) \\ \textrm{Eq. (1):}\qquad\quad 15 & = 3v_A+4v_B \\ \textrm{Eq. (1)':}\qquad\quad v_A & = 5-\frac{4}{3}v_B \\ \end{aligned}

\begin{aligned} \textrm{CoE:}\qquad\textrm{KE}_{\textrm{initial}} & = \textrm{KE}_{\textrm{final}} \\ \sum_{i=A}^{B}\frac{1}{2}m_iu_i^2 & = \sum_{i=A}^{B}\frac{1}{2}m_iv_i^2 \\ m_Au_A^2+m_Bu_B^2 & = m_Av_A^2+m_Bv_B^2 \\ (3)(5)^2 + (4)(0)^2 & = (3)(v_A)^2 + (4)(v_B)^2 \\ \textrm{Eq. (2):}\qquad\qquad 75 & = 3v_A^2 + 4v_B^2 \\ \end{aligned}

\begin{aligned} 75 & = 3\bigg( 5-\frac{4}{3}v_B\bigg)^2 + 4v_B^2 \\ 25 & = \bigg( 25 - \frac{40}{3}v_B + \frac{16}{9}v_B^2\bigg) + \frac{4}{3}v_B^2 \\ 0 & = v_B\bigg( -\frac{40}{3} + \frac{28}{9}v_B \bigg) \\ v_B & = 0\textrm{ \scriptsize{OR} }\frac{30}{7} \\ v_A & = \bigg[ 5-\frac{4}{3}(0)\bigg] \textrm{ \scriptsize{OR} }\bigg[ 5-\frac{4}{3}\bigg(\frac{30}{7}\bigg)\bigg] \\ & = 5\textrm{ \scriptsize{OR} }-\frac{5}{7} \\ \end{aligned}

(v_A,v_B) = \bigg( \displaystyle{-\frac{5}{7} , \frac{30}{7}}\bigg) \qquad \text{}^*\textrm{(5,0) is rejected.}

This contradicts with v_B-2.5=v_A>0. As the law of conservation of momentum is always observed, herein has energy \textrm{\scriptsize{NOT}} been conserved.

This problem is not to be attempted.