Three particles \(A\), \(B\) and \(C\) of masses 5 kg, 1 kg and 2 kg respectively lie at rest in that order on a straight smooth horizontal track \(XYZ\). Initially \(A\) is at \(X\), \(B\) is at \(Y\) and \(C\) is at \(Z\). Particle \(A\) is projected towards \(B\) with a speed of \(6\text{ ms}^{-1}\) and at the same instant \(C\) is projected towards \(B\) with a speed of \(v\text{ ms}^{-1}\). In the subsequent motion, \(A\) collides and coalesces with \(B\) to form particle \(D\). Particle \(D\) then collides and coalesces with \(C\) to form particle \(E\) and \(E\) moves towards \(Z\).
- Show that after the second collision the speed of \(E\) is \(\frac{15-v}{4}\text{ ms}^{-1}\). [3]
- The total loss of kinetic energy of the system due to the two collisions is 63 J.
Use the result from (a) to show that \(v = 3\). [3]
- It is given that the distance \(XY\) is 36 m and the distance \(YZ\) is 98 m.
- Find the time between the two collisions. [4]
- Find the time between the instant that \(A\) is projected from \(X\) and the instant that \(E\) reaches \(Z\). [1]