June 2023 Paper 1 Q12

12 In this question the unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal and vertically upwards respectively.

A particle has mass 2 kg.

(a) Write down its weight as a vector. [1]

A horizontal force of 3 N in the \(\mathbf{i}\) direction and a force \(\mathbf{F} = (-4\mathbf{i} + 12\mathbf{j})\) N act on the particle.

(b) Determine the acceleration of the particle. [3]

(c) The initial velocity of the particle is \(5\mathbf{i}\,\mathrm{m\,s^{-1}}\).
Find the velocity of the particle after 4 s. [2]

(d) Find the extra force that must be applied to the particle for it to move at constant velocity. [1]

Scheme	Marks	AO
Weight \(= -2g\mathbf{j}\) N \(\quad [= -19.6\mathbf{j}\text{ N}]\)	B1	2.5
	[1]

Mark scheme (b)

Scheme	Marks	AO
horizontal force \(3\mathbf{i}\)	B1	3.3
Newton’s second law \((-4\mathbf{i} + 12\mathbf{j}) + 3\mathbf{i} - 19.6\mathbf{j} = 2\mathbf{a}\)	M1	1.1a
\(\mathbf{a} = -0.5\mathbf{i} - 3.8\mathbf{j}\ \mathrm{m\,s^{-2}}\)	A1	1.1b
	[3]

B1: May be implied by correct resultant force

M1: Must be vector equation. allow one missing or incorrect force.

A1: ISW if the magnitude is also given

Scheme	Marks
Horizontal motion \(-4 + 3 = 2a_x\)	B1
Vertical motion \(12 - 19.6 = 2a_y\)	M1
\(\mathbf{a} = -0.5\mathbf{i} - 3.8\mathbf{j}\ \mathrm{m\,s^{-2}}\)	A1

B1: 3 N force used in the horizontal equation and not used in the vertical equation

M1: Considers motion in two directions. Allow one missing or incorrect force

A1: ISW if the magnitude is also given

Mark scheme (c)

Scheme	Marks	AO
\(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) \(= 5\mathbf{i} + (-0.5\mathbf{i} - 3.8\mathbf{j}) \times 4\)	M1	1.1a
\(= 3\mathbf{i} - 15.2\mathbf{j}\ \mathrm{m\,s^{-1}}\)	A1	1.1b
	[2]

M1: Using suvat equation(s) leading to a vector \(\mathbf{v}\). Do not award if scalar added to vector

A1: Mark final answer. Must be vector \(\mathbf{v}\) and not speed.
FT their vector acceleration