M1 June 2014 (R) Q5

EdexcelOld spec14 marksConnected Particles SUVAT

Figure 2: particles A (2m) and B (3m) hanging from a string over a fixed pulley — Figure 2

Two particles \(A\) and \(B\) have masses \(2m\) and \(3m\) respectively. The particles are connected by a light inextensible string which passes over a smooth light fixed pulley. The system is held at rest with the string taut. The hanging parts of the string are vertical and \(A\) and \(B\) are above a horizontal plane, as shown in Figure 2. The system is released from rest.

(a) Show that the tension in the string immediately after the particles are released is \(\dfrac{12}{5}mg\). (6)

After descending 1.5 m, \(B\) strikes the plane and is immediately brought to rest. In the subsequent motion, \(A\) does not reach the pulley.

(b) Find the distance travelled by \(A\) between the instant when \(B\) strikes the plane and the instant when the string next becomes taut. (6)

Given that \(m = 0.5\) kg,

Mark scheme (a)

Scheme	Marks
\(3mg - T = 3ma\)	M1A1
\(T - 2mg = 2ma\)	M1A1
\(T = 2mg + 2\left(mg - \dfrac{T}{3}\right)\)	DM1
\(T = \dfrac{12}{5}mg\) *Given Answer*	A1
	(6)

Notes

First M1 for resolving vertically (up or down) for \(B\), with correct no. of terms etc (allow if they omit \(m\) but have the 3)

First A1 for a correct equation.

Second M1 for resolving vertically (up or down) for \(A\), with correct no. of terms etc (allow if they omit \(m\) but have the 2)

Second A1 for a correct equation

Third M1, dependent on the first two M marks, for eliminating \(a\)

Third A1 for \(T = 12mg/5\) given answer

N.B. Either equation above can be replaced by the whole system equation

M1A1 for \(3mg - 2mg = 5ma\); any error loses both marks.

N.B. If \(m\) has been omitted in (a), which has led to a dimensionally incorrect value of \(a\), can score max B0M1A0M1M1A0 in (b) and M1A0 in (c).

Mark scheme (b)

Scheme	Marks
\(a = \dfrac{g}{5}\)	B1
At time of impact \(v^2 = u^2 + 2as = 2 \times \dfrac{g}{5} \times 1.5 = 0.6g\)	M1A1
Vertical motion under gravity \(0 = 0.6g - 2gs\)	M1
\(s = 0.3\) (m)
Total distance \(2 \times 0.3 = 0.6\) (m)	DM1A1
	(6)

Notes

B1 for \(a = g/5\) found (possibly in part (a)) and used here.

First M1 for using suvat with their \(a\) from part (a), to find the speed \(v\) (or \(v^2\)) of \(B\) at impact

First A1 for \(\sqrt{0.6g}\) oe, 2.4 or better (may be implied) found correctly.

Second M1 for using suvat with \(a = \pm g\), to obtain an equation in \(s\) only, using their \(v\) (or \(v^2\)) with final velocity \(= 0\)

Third M1, dependent on second M1, for doubling their \(s\) value

Second A1 for 0.6 (m)

Mark scheme (c)

Scheme	Marks
Impulse \(= 3m(v - u) = -3mu\)	M1
Magnitude \(= 3m\sqrt{0.6g} = 3.6\) (Ns) (3.64)	A1
	(2)
	(14 marks)

Notes

M1 for \(\pm 3m \times\) (their \(v\)) or \(\pm 1.5 \times\) (their \(v\)) or \(\pm m \times\) (their \(v\)) or \(\pm 0.5 \times\) (their \(v\))

M0 if \(3m\) missing or extra \(g\)

A1 for 3.6 or 3.64 (Ns)