M1 June 2015 Q5

EdexcelOld spec12 marksMoments

Figure 3: horizontal beam AB of length 5 m held by vertical ropes at A and at C, with CB = 0.5 m — Figure 3

A beam \(AB\) has length 5 m and mass 25 kg. The beam is suspended in equilibrium in a horizontal position by two vertical ropes. One rope is attached to the beam at \(A\) and the other rope is attached to the point \(C\) on the beam where \(CB = 0.5\) m, as shown in Figure 3. A particle \(P\) of mass 60 kg is attached to the beam at \(B\) and the beam remains in equilibrium in a horizontal position. The beam is modelled as a uniform rod and the ropes are modelled as light strings.

(a) Find

(i) the tension in the rope attached to the beam at \(A\),

(ii) the tension in the rope attached to the beam at \(C\). (6)

Particle \(P\) is removed and replaced by a particle \(Q\) of mass \(M\) kg at \(B\). Given that the beam remains in equilibrium in a horizontal position,

(b) find

(i) the greatest possible value of \(M\),

(ii) the greatest possible tension in the rope attached to the beam at \(C\). (6)

Mark scheme (a)

Scheme	Marks
\(T_A + T_C = 85g\)
OR \(M(A),\ 25g \times 2.5 + 60g \times 5 = 4.5 \times T_C\)	M1 A1
OR \(M(C),\ T_A \times 4.5 + 60g \times 0.5 = 25g \times 2\)
OR \(M(B),\ T_A \times 5 + T_C \times 0.5 = 25g \times 2.5\)
OR \(M(G),\ T_A \times 2.5 + 60g \times 2.5 = 2 \times T_C\)	M1 A1
\(T_A = \dfrac{40g}{9} = 44\) N or 43.6 N; \(\ T_C = \dfrac{725g}{9} = 790\) N or 789 N	A1; A1
	(6)

Notes

First M1 for a moments or vertical resolution equation, with correct no. of terms and dimensionally correct.

First A1 for a correct equation.

Second M1 for a moments equation, with correct no. of terms and dimensionally correct.

Second A1 for a correct equation.

Third A1 for 44 (N) or 43.6 (N) or \(40g/9\)

Fourth A1 for 790 (N) or 789 (N) or \(725g/9\)

Deduct 1 mark for inexact multiples of \(g\)

N.B. If they assume that both tensions are the same, can only score max M1 in (a) for \(M(A)\) or \(M(C)\).

If a vertical resolution is used, please give marks for this equation FIRST. If not, enter marks for each moments equation in the order in which they appear.

Mark scheme (b)

Scheme	Marks
\(\mathrm{M}(C),\ 25g \times 2 = Mg \times 0.5\)	M1 A1
(i) \(M = 100\)	A1
(ii) \(T_C = 25g + 100g\)	M1 A1
\(T_C = 125g\) (1200 or 1230) N	B1
	(6)
	(12 marks)

Notes

SCHEME CHANGE

B1 BECOMES THE FOURTH A1

First M1 for a moments equation with \(T_A = 0\)

First A1 for a correct equation

Second A1 for \(M = 100\)

Second M1 for a(nother) moments or vertical resolution equation with \(T_A = 0\)

Third A1 for a correct equation

Fourth A1 (B1) for \(T_C = 125g\) or 1230 (N) or 1200 (N)

N.B. Some candidates may need to solve 2 simult. equations in \(M\) and \(T_C\) and so will earn the ‘equation’ marks before they earn Second and Fourth A (B) marks.

If a vertical resolution is used, please give marks for this equation SECOND. If not, enter marks for each moments equation in the order in which they appear.

The possible equations are:

\(T_C = 25g + Mg\)

\(\mathrm{M}(C),\ 25g \times 2 = Mg \times 0.5\)

\(\mathrm{M}(A),\ 25g \times 2.5 + 5Mg = 4.5\,T_C\)

\(\mathrm{M}(B),\ 25g \times 2.5 = T_C \times 0.5\)

\(\mathrm{M}(G),\ T_C \times 2 = Mg \times 2.5\)

Any two of these can each earn M1A1 (M0 if incorrect no. of terms)

Then Second A1 for \(M = 100\)

And Fourth A1 (B1) for \(T_C = 125g\) or 1230 or 1200

N.B. No marks in (b) if they use any answers from (a) or \(M = 60\)