October 2021 Paper 1 Q13

OCR MEICurrent spec13 marksVariable Acceleration (Calculus)Vectors

13 In this question \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors in the \(x\)- and \(y\)-directions respectively.

The velocity of a particle at time \(t\) s is given by \((3t^2\mathbf{i} + 7\mathbf{j})\,\text{m}\,\text{s}^{-1}\). At time \(t = 0\) the position of the particle with respect to the origin is \((-\mathbf{i} + 2\mathbf{j})\) m.

(a) Determine the distance of the particle from the origin when \(t = 2\). [6]

(b) Show that the cartesian equation of the path of the particle is \(x = \left(\dfrac{y-2}{7}\right)^3 - 1\). [3]

(c) At time \(t = 2\), the magnitude of the resultant force acting on the particle is 48 N.

Find the mass of the particle. [4]

Mark scheme (a)

Scheme	Marks	AO
\(\mathbf{r} = \displaystyle\int(\mathbf{v})\,\mathrm{d}t = \int\left(3t^2\mathbf{i} + 7\mathbf{j}\right)\mathrm{d}t\) \(= t^3\mathbf{i} + 7t\mathbf{j} + \mathbf{c}\)	M1*	3.4
When \(t = 0,\ \mathbf{r} = -\mathbf{i} + 2\mathbf{j} = \mathbf{c}\) So \(\mathbf{r} = (t^3 - 1)\mathbf{i} + (7t + 2)\mathbf{j}\)	M1 dep*	1.1b
When \(t = 2\), \(\mathbf{r} = (2^3 - 1)\mathbf{i} + (7 \times 2 + 2)\mathbf{j} = 7\mathbf{i} + 16\mathbf{j}\)	M1 dep* A1	3.1b 1.1b
distance \(= \sqrt{7^2 + 16^2} = \sqrt{305}\)	M1	3.1b
distance = 17.5 m	A1	3.2a
	[6]

Notes

M1*: Attempt to integrate both components; condone missing \(+\mathbf{c}\)

M1 dep*: Using initial conditions

M1 dep*: Using \(t = 2\) to find position vector or values for \(x\) and \(y\)

A1: Accept vector form or two clear components.

M1: Using Pythagoras

A1: FT their components

Mark scheme (b)

Scheme	Marks	AO
using \(x = t^3 - 1,\ y = 7t + 2\)	M1	3.1a
Substitute \(t = \dfrac{y-2}{7}\) into equation for \(x\)	M1 dep	1.1b
\(x = \left(\dfrac{y-2}{7}\right)^3 - 1\) AG	A1	1.1b
	[3]

Notes

M1: Extracting equations for \(x\) and \(y\) from their displacement vector

M1 dep: Attempt to eliminate \(t\)

A1: cao

Equivalent form \(y = 7(x+1)^{\frac{1}{3}} + 2\) for M1M1A0

Mark scheme (c)

Scheme	Marks	AO
\(\mathbf{a} = \dfrac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} = 6t\mathbf{i}\)	M1*	3.1b
When \(t = 2,\ \mathbf{a} = 12\mathbf{i}\)	M1*	3.4
The force must be in that direction, so \(\mathbf{F} = 48\mathbf{i} = m\mathbf{a}\)	M1*	3.1b
So \(m = 4\) kg	A1 dep*	1.1b
	[4]

Notes

M1*: Must be vector acceleration

M1*: Evaluating when \(t = 2\)
\(a = 12\) is sufficient here

M1*: Newton’s second law in vector form, or in \(x\)-direction only
If their \(\mathbf{a}\) has two non-zero components, allow for dividing 48 by the magnitude of their \(\mathbf{a}\)

A1 dep*: cao