Problems

Filters
Clear Filters

2 problems found

2012 Paper 3 Q11
D: 1700.0 B: 1500.0
energy conservation variable mass falling chain rope acceleration differentiation kinetic energy potential energy gravity

One end of a thin heavy uniform inextensible perfectly flexible rope of length \(2L\) and mass \(2M\) is attached to a fixed point \(P\). A particle of mass \(m\) is attached to the other end. Initially, the particle is held at \(P\) and the rope hangs vertically in a loop below \(P\). The particle is then released so that it and a section of the rope (of decreasing length) fall vertically as shown in the diagram.

\psset{xunit=1.0cm,yunit=0.9cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.3pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(0.13,-0.26)(3.26,5.51) \psline(1,5)(3,5) \psline[linewidth=0.1pt,linestyle=dashed,dash=2pt 2pt]{<->}(1.52,0)(1.52,5) \psline[linewidth=0.1pt,linestyle=dashed,dash=2pt 2pt]{<->}(2.53,3.2)(2.53,5) \psline(2.1,3.18)(2.06,0.25) \psline(2,5)(2.02,0.26) \psline(2.02,0.26)(2.03,0) \psline(2.03,0)(2.06,0.25) \rput[tl](1.94,5.45){\(P\)} \rput[tl](2.6,4.25){\(x\)} \rput[tl](0.2,2.85){\(L+\frac{1}{2}x\)} \begin{scriptsize} \psdots[dotsize=4pt 0,dotstyle=*](2.1,3.18) \end{scriptsize} \end{pspicture*}
You may assume that each point on the moving section of the rope falls at the same speed as the particle. Given that energy is conserved, show that, when the particle has fallen a distance \(x\) (where \(x< 2L\)), its speed \(v\) is given by \[ v^2 = \frac { 2g x \big( mL +ML - \frac14 Mx)}{mL +ML - \frac12 Mx}\,. \] Hence show that the acceleration of the particle is \[ g + \frac{ Mgx\big(mL+ML- \frac14 Mx\big)}{2\big(mL +ML -\frac12 Mx\big)^2}\, \,.\] Deduce that the acceleration of the particle after it is released is greater than \(g\).

View
1999 Paper 3 Q10
D: 1700.0 B: 1484.0
momentum impulse inelastic collision chain problem summation to integral approximation falling chain variable mass

A chain of mass \(m\) and length \(l\) is composed of \(n\) small smooth links. It is suspended vertically over a horizontal table with its end just touching the table, and released so that it collapses inelastically onto the table. Calculate the change in momentum of the \((k+1)\)th link from the bottom of the chain as it falls onto the table. Write down an expression for the total impulse sustained by the table in this way from the whole chain. By approximating the sum by an integral, show that this total impulse is approximately \[ {\textstyle \frac23} m \surd(2gl) \] when \(n\) is large.

View