M1 June 2018 Q7

EdexcelOld spec16 marksConnected Particles Resolving Forces

Figure 4: particle P (4m) at X on a rough plane inclined at alpha, a distance h from the bottom, joined by a string over a pulley at the top to particle Q (m) hanging a distance d below the pulley — Figure 4

A particle \(P\) of mass \(4m\) is held at rest at the point \(X\) on the surface of a rough inclined plane which is fixed to horizontal ground. The point \(X\) is a distance \(h\) from the bottom of the inclined plane. The plane is inclined to the horizontal at an angle \(\alpha\) where \(\tan\alpha = \dfrac{3}{4}\). The coefficient of friction between \(P\) and the plane is \(\dfrac{1}{4}\). The particle \(P\) is attached to one end of a light inextensible string. The string passes over a small smooth pulley which is fixed at the top of the plane. The other end of the string is attached to a particle \(Q\) of mass \(m\) which hangs freely at a distance \(d\), where \(d > h\), below the pulley, as shown in Figure 4.

The string lies in a vertical plane through a line of greatest slope of the inclined plane. The system is released from rest with the string taut and \(P\) moves down the plane.

For the motion of the particles before \(P\) hits the ground,

(a) state which of the information given above implies that the magnitudes of the accelerations of the two particles are the same, (1)

(b) write down an equation of motion for each particle, (5)

When \(P\) hits the ground, it immediately comes to rest. Given that \(Q\) comes to instantaneous rest before reaching the pulley,

(d) show that \(d > \dfrac{28h}{25}\). (5)

Scheme	Marks
Inextensible string	B1
	(1)

Mark scheme (b)

Scheme	Marks
\(4mg\sin\alpha - T - F = 4ma\)	M1 A2
\(T - mg = ma\)	M1 A1
	(5)

Notes

MARK PARTS (b) and (c) together

N.B. Omission of \(m\) is a Method error i.e. M0 for that equation

First M1 for equation of motion for \(P\) with usual rules (omission of 4 on RHS is M0)

First A1 and second A1 for a correct equation, A1A0 if one error

Second M1 for equation of motion for \(Q\) with usual rules

Third A1 for a correct equation

Use of e.g cos(4/5) instead of \(\cos\alpha\) is an A error unless they recover correctly.

N.B. Allow consistent use of \(-a\)

Mark scheme (c)

Scheme	Marks
\(F = \tfrac{1}{4}R\)	B1
\(R = 4mg\cos\alpha\)	B1
\(\cos\alpha = \tfrac{4}{5}\) or \(\sin\alpha = \tfrac{3}{5}\)	B1
Eliminating \(R\), \(F\) and \(T\)	M1
\(a = \tfrac{3}{25}g = 1.2\) or 1.18 (m s\(^{-2}\))	A1
	(5)

Notes

MARK PARTS (b) and (c) together

First B1 for \(F = \tfrac{1}{4}R\) seen or implied

Second B1 for \(R = 4mg\cos\alpha\) seen or implied

Third B1 for \(\cos\alpha = \tfrac{4}{5}\) or \(\sin\alpha = \tfrac{3}{5}\) seen or implied or an appropriate correct angle is used to give a correct trig ratio

First M1 for eliminating \(R\), \(F\) and \(T\) and finding an \(a\) value

First A1 \(a = \tfrac{3}{25}g = 1.2\) or 1.18 (m s\(^{-2}\)) (must be positive)

Mark scheme (d)

Scheme	Marks
\(v^2 = 2 \times \tfrac{3}{25}gh = \tfrac{6}{25}gh\)	M1
\(0^2 = \tfrac{6}{25}gh - 2gs\)
\(s = \tfrac{3}{25}h\)	M1 A1
\(d > \tfrac{3}{25}h + h = \tfrac{28}{25}h\) GIVEN ANSWER	DM1 A1
	(5)
	(16 marks)

Notes

First M1 for finding \(v\) or \(v^2\) for \(P\) using their \(a\) (M0 if \(g\) is used)

Second M1 for a complete method to find \(s\), independent but must have found \(v\) or \(v^2\) (M0 if \(g\) not used)

First A1 for \(s = \tfrac{3}{25}h\) oe

Third DM1, dependent on previous two M’s, for adding \(h\) onto their \(s\) oe

Second A1 for GIVEN ANSWER