Problems

Filters
Clear Filters

3 problems found

2006 Paper 2 Q10
D: 1600.0 B: 1500.0
momentum collisions coefficient of restitution Newton's law of restitution conservation of momentum inequality three particles successive collisions

Three particles, \(A\), \(B\) and \(C\), of masses \(m\), \(km\) and \(3m\) respectively, are initially at rest lying in a straight line on a smooth horizontal surface. Then \(A\) is projected towards \(B\) at speed \(u\). After the collision, \(B\) collides with \(C\). The coefficient of restitution between \(A\) and \(B\) is \(\frac12\) and the coefficient of restitution between \(B\) and \(C\) is \(\frac14\).

  1. Find the range of values of \(k\) for which \(A\) and \(B\) collide for a second time.
  2. Given that \(k=1\) and that \(B\) and \(C\) are initially a distance \(d\) apart, show that the time that elapses between the two collisions of \(A\) and \(B\) is \(\dfrac{60d}{13u}\,\).

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 \(\frac{128d}{3 \times 13u} + \frac{4d}{3u} = \frac{(128+52)d}{3 \times 13 u} = \frac{60d}{13u}\)

View
1998 Paper 1 Q9
D: 1500.0 B: 1484.0
SHM simple pendulum Newton's experimental law coefficient of restitution collision conservation of momentum successive collisions equal pendulums period

Two small spheres \(A\) and \(B\) of equal mass \(m\) are suspended in contact by two light inextensible strings of equal length so that the strings are vertical and the line of centres is horizontal. The coefficient of restitution between the spheres is \(e\). The sphere \(A\) is drawn aside through a very small distance in the plane of the strings and allowed to fall back and collide with the other sphere \(B\), its speed on impact being \(u\). Explain briefly why the succeeding collisions will all occur at the lowest point. (Hint: Consider the periods of the two pendulums involved.) Show that the speed of sphere \(A\) immediately after the second impact is \(\frac{1}{2}u(1+e^{2})\) and find the speed, then, of sphere \(B\).

View
1992 Paper 3 Q11
D: 1700.0 B: 1500.0
momentum and collisions coefficient of restitution relative velocity kinematics constant acceleration particle barrier successive collisions

\(\,\)

TikZ diagram
\(AOB\) represents a smooth vertical wall and \(XY\) represents a parallel smooth vertical barrier, both standing on a smooth horizontal table. A particle \(P\) is projected along the table from \(O\) with speed \(V\) in a direction perpendicular to the wall. At the time of projection, the distance between the wall and the barrier is \((75/32)VT\), where \(T\) is a constant. The barrier moves directly towards the wall, remaining parallel to the wall, with initial speed \(4V\) and with constant acceleration \(4V/T\) directly away from the wall. The particle strikes the barrier \(XY\) and rebounds. Show that this impact takes place at time \(5T/8\). The barrier is sufficiently massive for its motion to be unaffected by the impact. Given that the coefficient of restitution is \(1/2\), find the speed of \(P\) immediately after impact. \(P\) strikes \(AB\) and rebounds. Given that the coefficient of restitution for this collision is also \(1/2,\) show that the next collision of \(P\) with the barrier is at time \(9T/8\) from the start of the motion.

View