M5 June 2017 Q6

EdexcelOld spec12 marksImpulse & Momentum

6. A small object \(P\), of mass \(m_0\), is projected vertically upwards from the ground with speed \(U\). As \(P\) moves upwards it picks up droplets of moisture from the atmosphere. The droplets are at rest immediately before they are picked up. In a model of the motion, \(P\) is modelled as a particle, air resistance is assumed to be negligible and the acceleration due to gravity is assumed to have the constant value of \(g\). When \(P\) is at a height \(x\) above the ground, the combined mass of \(P\) and the moisture is \(m_0(1 + kx)\), where \(k\) is a constant, and the speed of \(P\) is \(v\).

(a) Show that, while \(P\) is moving upwards \[\frac{\mathrm{d}}{\mathrm{d}x}\left(v^2\right) + \frac{2kv^2}{(1 + kx)} = -2g\] (7)

The general solution of this differential equation is given by \(v^2 = \dfrac{A}{(1 + kx)^2} - \dfrac{2g}{3k}(1 + kx)\), where \(A\) is an arbitrary constant.

Given that \(U = \sqrt{2gh}\) and \(k = \dfrac{7}{3h}\)

(b) find, in terms of \(h\), the height of \(P\) above the ground when \(P\) first comes to rest. (5)

Mark scheme (a)

Scheme	Marks
\(m = m_0(1 + kx) \qquad \dfrac{\mathrm{d}m}{\mathrm{d}t} = m_0kv\)	M1 A1
\((m + \delta m)(v + \delta v) - mv = -mg\,\delta t\)	M1
\(\dfrac{\mathrm{d}v}{\mathrm{d}t} + \dfrac{v}{m}\dfrac{\mathrm{d}m}{\mathrm{d}t} = -g\)	A1
\(\dfrac{\mathrm{d}v}{\mathrm{d}t} + \dfrac{vm_0kv}{m_0(1 + kx)} = -g\)	M1
\(\dfrac{v\,\mathrm{d}v}{\mathrm{d}x} + \dfrac{kv^2}{(1 + kx)} = -g\)	M1
\(\dfrac{\mathrm{d}(v^2)}{\mathrm{d}x} + \dfrac{2kv^2}{(1 + kx)} = -2g\) printed answer	A1
	(7)

Notes

First M1 for differentiating \(m\) wrt to \(t\), and using \(v = \dfrac{\mathrm{d}x}{\mathrm{d}t}\)

First A1 for correct expression for \(\dfrac{\mathrm{d}m}{\mathrm{d}t}\) in terms of \(v\)

Second M1 for impulse-momentum principle

Second A1 for a correct diff equn in \(m\), \(v\) and \(t\).

Third M1 for sub for \(m\) and \(\dfrac{\mathrm{d}m}{\mathrm{d}t}\)

Fourth M1 for changing \(\dfrac{\mathrm{d}v}{\mathrm{d}t}\) to \(v\dfrac{\mathrm{d}v}{\mathrm{d}x}\) to \(\dfrac{1}{2}\dfrac{\mathrm{d}(v^2)}{\mathrm{d}x}\)

Third A1 for PRINTED ANSWER

Mark scheme (b)

Scheme	Marks
\(x = 0,\ v^2 = 2gh \Rightarrow 2gh = A - \dfrac{2g}{3k} \Rightarrow A = 2gh + \dfrac{2g}{3k}\)	M1 A1
\(v = 0 \Rightarrow \dfrac{3kA}{2g} = (1 + kH)^3\)	M1
\(3kh + 1 = (1 + kH)^3\)
\(\dfrac{3h}{7} = H\)	M1 A1
	(5)
	(12 marks)

Notes

First M1 for using initial conditions

First A1 for a correct expression for \(A\)

Second M1 for putting \(v = 0\)

Third M1 for solving for \(H\) in terms of \(h\) only.

Second A1 for \(3h/7\).