FM1-1 Momentum and impulse — revision question pack

2 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section FM1-1. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

FM1-1.1 · Momentum and impulse. The impulse-momentum principle. The principle of conservation of momentum applied to two spheres colliding directly.

Explanation

For a particle moving along a line, momentum is the signed quantity $mv$ . A positive direction must therefore be fixed before velocities enter an equation.
Impulse is the change in momentum, $I=m(v-u)$ , and also $I=\int F\,dt$ , the signed area under a force-time graph.
During a direct collision of two spheres modelled as particles, total momentum is conserved when the external impulse over the short impact is negligible.
Write one momentum equation for the system and separate impulse equations for individual spheres when required.
Examiners expect signed velocities throughout; a negative result describes motion opposite to the chosen direction and must not be silently changed into a positive speed.

A direct collision drawn on one signed line, with the chosen positive direction shown.

Worked example

Sphere $A$ of mass $2\,\text{kg}$ moves right at $5\,\text{m s}^{-1}$ and collides directly with sphere $B$ of mass $3\,\text{kg}$ moving left at $1\,\text{m s}^{-1}$ . They coalesce. Find the common velocity and the impulse on $A$ .

1.Take rightwards as positive and conserve momentum: $2(5)+3(-1)=5v$ .
2.Thus $v=\dfrac75\,\text{m s}^{-1}$ .
3.For $A$ , $I=2(\tfrac75-5)=-\dfrac{36}{5}\,\text{N s}$ .

Answer: The common velocity is $\dfrac75\,\text{m s}^{-1}$ rightwards; the impulse on $A$ is $\dfrac{36}{5}\,\text{N s}$ leftwards.

Common mistakes

Don't use positive speeds for both spheres even though one initial velocity is in the negative direction.
Don't calculate impulse as $m(u-v)$ instead of the final momentum minus the initial momentum.
Don't conserve kinetic energy when the spheres coalesce.

Exam tip

State the positive direction above the first momentum equation and interpret every negative final value against it.

Tier 1 · Easy

1.
Rightwards is positive. A $3\,\text{kg}$ particle changes velocity from $4\,\text{m s}^{-1}$ rightwards to $2\,\text{m s}^{-1}$ leftwards. Find its impulse.
(2)
(Total for Question 1 is 2 marks)
2.
Rightwards is positive. A $2\,\text{kg}$ particle $A$ moving at $4\,\text{m s}^{-1}$ collides directly with a $3\,\text{kg}$ particle $B$ moving at $-1\,\text{m s}^{-1}$ . Immediately afterwards, $A$ has velocity $-0.5\,\text{m s}^{-1}$ . Find the velocity of $B$ and the speed at which the particles are separating.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Two spheres collide head-on and remain together. The first has mass $2\,\text{kg}$ and velocity $7\,\text{m s}^{-1}$ ; the second has mass $3\,\text{kg}$ and velocity $-1\,\text{m s}^{-1}$ . Calculate their common velocity.
(3)
(Total for Question 1 is 3 marks)
2.
Rightwards is positive. A $3\,\text{kg}$ particle $P$ moves rightwards at $4\,\text{m s}^{-1}$ . A $2\,\text{kg}$ particle $Q$ has speed $3\,\text{m s}^{-1}$ , but its initial direction is unknown. After a direct collision, $P$ is at rest and $Q$ moves rightwards at $3\,\text{m s}^{-1}$ . Determine $Q$ 's initial direction and the impulse on each particle.
(4)
(Total for Question 2 is 4 marks)
3.
Rightwards is positive. A particle of mass $5\,\text{kg}$ is moving leftwards at $2\,\text{m s}^{-1}$ . A constant force of magnitude $28\,\text{N}$ acts rightwards for $0.5\,\text{s}$ . Find the impulse of the force and the particle's final velocity, stating whether its direction of motion is reversed.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
Sphere $A$ , of mass $m\,\text{kg}$ , moves rightwards at $8\,\text{m s}^{-1}$ behind a $2\,\text{kg}$ sphere $B$ moving rightwards at $2\,\text{m s}^{-1}$ . After their direct collision, the velocities of $A$ and $B$ are respectively $1\,\text{m s}^{-1}$ and $5\,\text{m s}^{-1}$ rightwards. Find $m$ and the impulse on $A$ .
(4)
(Total for Question 1 is 4 marks)
2.
Rightwards is positive. A $2\,\text{kg}$ particle $P$ moving at $6\,\text{m s}^{-1}$ collides directly with a $4\,\text{kg}$ particle $Q$ moving at $-1\,\text{m s}^{-1}$ . The impulse on $Q$ is $J\,\text{N s}$ rightwards. Determine the complete range of possible values of $J$ for which both particles move rightwards after impact and separate with $Q$ ahead.
(5)
(Total for Question 2 is 5 marks)
3.
Rightwards is positive. Sphere $A$ , of mass $3\,\text{kg}$ , moves rightwards at $7\,\text{m s}^{-1}$ and collides directly with sphere $B$ , of mass $5\,\text{kg}$ , moving leftwards at $1\,\text{m s}^{-1}$ . Immediately after the collision, $A$ has velocity $-2\,\text{m s}^{-1}$ . The contact lasts $0.015\,\text{s}$ . Find the velocity of $B$ , the impulse on each sphere and the average force exerted on $A$ . Verify that the spheres separate after impact, with $B$ ahead of $A$ .
(6)
(Total for Question 3 is 6 marks)
4.
Rightwards is positive. A sphere $A$ of mass $4\,\text{kg}$ initially moves at $1\,\text{m s}^{-1}$ . A force-time graph is triangular: the force rises uniformly from zero to $H\,\text{N}$ in $2\,\text{s}$ and then falls uniformly to zero in a further $3\,\text{s}$ . The force acts rightwards and leaves $A$ moving at $6\,\text{m s}^{-1}$ . Sphere $A$ then collides directly with a sphere $B$ of unknown mass which is moving leftwards at $2\,\text{m s}^{-1}$ . During the collision, the impulse on $A$ is $12\,\text{N s}$ leftwards. Immediately after the collision, $B$ moves rightwards at $4\,\text{m s}^{-1}$ . Determine $H$ , $A$ 's post-collision velocity and the mass ratio $A:B$ . Verify that total momentum is conserved in the collision.
(7)
(Total for Question 4 is 7 marks)

FM1-1.2 · Momentum as a vector. The impulse-momentum principle in vector form.

Explanation

Momentum is a vector: $\mathbf p=m\mathbf v$ . The impulse-momentum principle is $\mathbf J=m(\mathbf v-\mathbf u)$ , so $\mathbf v=\mathbf u+\mathbf J/m$ .
Work component by component in a fixed orthogonal basis, retaining signs. Only after the new velocity or impulse vector is known should its magnitude and direction be calculated using Pythagoras and an angle with a quadrant check.
A stated impulse magnitude alone does not determine its components; direction information or another condition is needed.
Examiners expect vector units, a complete $\mathbf i,\mathbf j$ answer and any requested scalar speed distinguished from the velocity vector.
Perpendicularity is tested by a zero scalar product.

Vector impulse added to initial momentum to produce the final momentum.

Worked example

A $5\,\text{kg}$ particle initially moves with velocity $(2\mathbf i+3\mathbf j)\,\text{m s}^{-1}$ and receives impulse $(15\mathbf i-20\mathbf j)\,\text{N s}$ . Find its final velocity and speed.

1. $\mathbf v=\mathbf u+\mathbf J/m=(2\mathbf i+3\mathbf j)+(3\mathbf i-4\mathbf j)$ .
2.Therefore $\mathbf v=(5\mathbf i-\mathbf j)\,\text{m s}^{-1}$ .
3.The speed is $|\mathbf v|=\sqrt{5^2+(-1)^2}=\sqrt{26}\,\text{m s}^{-1}$ .

Answer: Final velocity $(5\mathbf i-\mathbf j)\,\text{m s}^{-1}$ ; speed $\sqrt{26}\,\text{m s}^{-1}$ .

Common mistakes

Don't add the impulse vector directly to velocity without dividing by the mass.
Don't use the impulse magnitude as both its $\mathbf i$ and $\mathbf j$ components.
Don't report $\sqrt{26}\,\text{m s}^{-1}$ as a velocity rather than a speed.

Exam tip

Keep the calculation in vector form until the final line, then calculate magnitude or direction only if requested.

Tier 1 · Easy

1.
A $2\,\text{kg}$ particle changes velocity from $(3\mathbf i-4\mathbf j)\,\text{m s}^{-1}$ to $(5\mathbf i+\mathbf j)\,\text{m s}^{-1}$ . Find the impulse vector and its magnitude.
(3)
(Total for Question 1 is 3 marks)
2.
A particle of mass $2\,\text{kg}$ has initial velocity $(3\mathbf i+\mathbf j)\,\text{m s}^{-1}$ . It receives successive impulses $(2\mathbf i-6\mathbf j)\,\text{N s}$ and $(-4\mathbf i+2\mathbf j)\,\text{N s}$ . Find its final velocity and speed.
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
A $4\,\text{kg}$ particle initially has velocity $(2\mathbf i-\mathbf j)\,\text{m s}^{-1}$ . It receives an impulse $(-6\mathbf i+8\mathbf j)\,\text{N s}$ . Determine its new speed and the acute angle its velocity makes above the positive $\mathbf i$ direction.
(4)
(Total for Question 1 is 4 marks)
2.
A $2\,\text{kg}$ particle initially has velocity $(\mathbf i-2\mathbf j)\,\text{m s}^{-1}$ . It receives an impulse whose $\mathbf i$ component is $4\,\text{N s}$ and whose $\mathbf j$ component is positive. Its final speed is $5\,\text{m s}^{-1}$ . Determine the final velocity and the impulse direction.
(5)
(Total for Question 2 is 5 marks)
3.
A particle of mass $m\,\text{kg}$ has velocity $(5\mathbf i-2\mathbf j)\,\text{m s}^{-1}$ . An impulse of $(-12\mathbf i+9\mathbf j)\,\text{N s}$ changes its velocity to $(\mathbf i+\mathbf j)\,\text{m s}^{-1}$ . Find $m$ and the speed of the particle after the impulse.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A $3\,\text{kg}$ particle has initial velocity $(4\mathbf i-2\mathbf j)\,\text{m s}^{-1}$ . An impulse of magnitude $15\,\text{N s}$ leaves its final velocity perpendicular to its initial velocity. Find all possible final velocities and the corresponding impulse vectors.
(5)
(Total for Question 1 is 5 marks)
2.
Particles $P$ and $Q$ , of masses $1\,\text{kg}$ and $2\,\text{kg}$ , each initially have velocity $(2\mathbf i+\mathbf j)\,\text{m s}^{-1}$ . The same impulse $\mathbf J=(a\mathbf i+b\mathbf j)\,\text{N s}$ is applied separately to each particle. Their resulting speeds are $5\sqrt2\,\text{m s}^{-1}$ and $\sqrt{85}/2\,\text{m s}^{-1}$ respectively. Given $b>0$ , determine the impulse and its direction.
(7)
(Total for Question 2 is 7 marks)
3.
A $2\,\text{kg}$ particle initially has velocity $(4\mathbf i-\mathbf j)\,\text{m s}^{-1}$ . A constant force $(-6\mathbf i+8\mathbf j)\,\text{N}$ acts for $t\,\text{s}$ , where $t>0$ . Immediately afterwards the velocity is parallel to $\mathbf i+2\mathbf j$ and has a positive $\mathbf i$ component. Determine $t$ , the final velocity and speed, and the impulse exerted on the particle.
(7)
(Total for Question 3 is 7 marks)
4.
A $4\,\text{kg}$ particle initially has velocity $(2\mathbf i-3\mathbf j)\,\text{m s}^{-1}$ . It receives successive impulses $\mathbf J_1=(a\mathbf i+12\mathbf j)\,\text{N s}$ and $\mathbf J_2=(-8\mathbf i+b\mathbf j)\,\text{N s}$ , where $a>0$ and $b>0$ . The impulse $\mathbf J_1$ is parallel to $3\mathbf i+4\mathbf j$ , and the two impulses are perpendicular. Find both impulses, the final velocity and the final speed.
(7)
(Total for Question 4 is 7 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

FM1-1.1 · Momentum and impulse. The impulse-momentum principle. The principle of conservation of momentum applied to two spheres colliding directly.

Tier 1 · Easy

Mark scheme for FM1-1.1 Tier 1 · Easy
Question	Scheme	Marks
1	$-18\,\text{N s}$ , so the impulse is $18\,\text{N s}$ leftwards	2
(2 marks)		2
Notes
Use $I=m(v-u)$ with $u=4$ and $v=-2$ . Hence $I=3(-2-4)=-18\,\text{N s}$ . The negative sign gives the leftward direction.
2	The initial momentum is $2(4)+3(-1)=5\,\text{kg m s}^{-1}$ Conservation of momentum gives $5=2(-0.5)+3v_B$ , so $v_B=2\,\text{m s}^{-1}$ The separation speed is $v_B-v_A=2-(-0.5)=2.5\,\text{m s}^{-1}$	3
(3 marks)		3
Notes
Initial momentum is $2(4)+3(-1)=5\,\text{kg m s}^{-1}$ . Conservation of momentum gives $5=2(-0.5)+3v_B$ , so $v_B=2\,\text{m s}^{-1}$ . Since $B$ is ahead and $v_B-v_A=2-(-0.5)=2.5>0$ , their distance apart increases after the collision.

Tier 2 · Standard

Mark scheme for FM1-1.1 Tier 2 · Standard
Question	Scheme	Marks
1	$\dfrac{11}{5}\,\text{m s}^{-1}$ in the positive direction	3
(3 marks)		3
Notes
Conservation of momentum gives $2(7)+3(-1)=(2+3)v$ . Thus $11=5v$ , so $v=11/5\,\text{m s}^{-1}$ in the positive direction.
2	The final momentum is $3(0)+2(3)=6\,\text{kg m s}^{-1}$ Conservation of momentum requires $u_Q=-3\,\text{m s}^{-1}$ , so $Q$ initially moves leftwards The impulse on $P$ is $3(0-4)=-12\,\text{N s}$ The impulse on $Q$ is $2[3-(-3)]=12\,\text{N s}$	4
(4 marks)		4
Notes
The final momentum is $3(0)+2(3)=6\,\text{kg m s}^{-1}$ . If $Q$ initially moved rightwards, the initial momentum would be $3(4)+2(3)=18$ , so conservation would fail. With $u_Q=-3$ , the initial momentum is $12-6=6$ , as required. The impulse on $P$ is $3(0-4)=-12\,\text{N s}$ , and the equal opposite impulse on $Q$ is $2(3-(-3))=12\,\text{N s}$ .
3	The impulse is $28(0.5)=14\,\text{N s}$ rightwards The change in velocity is $14/5\,\text{m s}^{-1}$ rightwards The final velocity is $-2+14/5=4/5\,\text{m s}^{-1}$ The final velocity is positive, so the particle's direction is reversed and it moves rightwards	4
(4 marks)		4
Notes
The constant force supplies impulse $I=F\Delta t=28(0.5)=14\,\text{N s}$ rightwards. Using $I=m(v-u)$ with $u=-2$ gives $14=5(v+2)$ , so $v=4/5\,\text{m s}^{-1}$ . Its positive sign shows that the particle has reversed direction.

Tier 3 · Hard

Mark scheme for FM1-1.1 Tier 3 · Hard
Question	Scheme	Marks
1	$m=\dfrac67\,\text{kg}$ Impulse on $A=-6\,\text{N s}$ , or $6\,\text{N s}$ leftwards	4
(4 marks)		4
Notes
Conservation of momentum gives $8m+2(2)=m(1)+2(5)$ . Therefore $7m=6$ and $m=6/7\,\text{kg}$ . For $A$ , $I=m(v-u)=(6/7)(1-8)=-6\,\text{N s}$ , so the impulse is leftwards.
2	The equal and opposite impulses give $v_P=6-J/2$ They give $v_Q=-1+J/4$ $P$ moves rightwards only if $J<12$ $Q$ moves rightwards only if $J>4$ Separation with $Q$ ahead requires $J>28/3$ , so the complete range is $28/3<J<12$	5
(5 marks)		5
Notes
Equal opposite impulses give $v_P=6-J/2$ and $v_Q=-1+J/4$ . Both are rightwards when $J<12$ and $J>4$ . Since $Q$ is ahead, separation requires $v_Q>v_P$ , so $-1+J/4>6-J/2$ , giving $J>28/3$ ; this is stronger than $J>4$ . Hence the complete interval satisfying all three signed-velocity conditions is $28/3<J<12$ .
3	The initial total momentum is $3(7)+5(-1)=16\,\text{kg m s}^{-1}$ Conservation of momentum gives $16=3(-2)+5v_B$ , so $v_B=22/5\,\text{m s}^{-1}$ The impulse on $A$ is $3(-2-7)=-27\,\text{N s}$ The impulse on $B$ is $+27\,\text{N s}$ The average force on $A$ is $-27/0.015=-1800\,\text{N}$ , or $1800\,\text{N}$ leftwards Since $22/5>-2$ and $B$ is ahead, the separation increases immediately after impact	6
(6 marks)		6
Notes
The initial momentum is $16\,\text{kg m s}^{-1}$ . Conservation of momentum gives $16=3(-2)+5v_B$ , hence $v_B=22/5\,\text{m s}^{-1}$ . The impulse on $A$ is $3[-2-7]=-27\,\text{N s}$ , so the impulse on $B$ is $+27\,\text{N s}$ . Dividing the impulse on $A$ by $0.015\,\text{s}$ gives an average force of $-1800\,\text{N}$ . As $B$ is ahead and $v_B-v_A=22/5-(-2)=32/5>0$ , the spheres separate.
4	The area of the force-time triangle is $\tfrac12(5)H=5H/2\,\text{N s}$ $4(6-1)=5H/2$ , so $H=8\,\text{N}$ For $A$ , $4(v_A-6)=-12$ Hence $v_A=3\,\text{m s}^{-1}$ rightwards The impulse on $B$ is $12\,\text{N s}$ rightwards If $B$ has mass $m\,\text{kg}$ , then $m[4-(-2)]=12$ , so $m=2$ and the mass ratio $A:B$ is $2:1$ The total momentum before is $4(6)+2(-2)=20\,\text{kg m s}^{-1}$ and after is $4(3)+2(4)=20\,\text{kg m s}^{-1}$	7
(7 marks)		7
Notes
The impulse from the force is the triangular area $5H/2$ . Equating this to $4(6-1)=20$ gives $H=8$ . The stated collision impulse gives $4(v_A-6)=-12$ , so $v_A=3\,\text{m s}^{-1}$ . Newton's third law gives $B$ the equal opposite impulse, so $m[4-(-2)]=12$ and $m=2\,\text{kg}$ . Thus the mass ratio is $4:2=2:1$ . Direct calculation gives total momentum $20\,\text{kg m s}^{-1}$ both before and after the collision.

FM1-1.2 · Momentum as a vector. The impulse-momentum principle in vector form.

Tier 1 · Easy

Mark scheme for FM1-1.2 Tier 1 · Easy
Question	Scheme	Marks
1	$(4\mathbf i+10\mathbf j)\,\text{N s}$ $2\sqrt{29}\,\text{N s}$	3
(3 marks)		3
Notes
$\mathbf J=m(\mathbf v-\mathbf u)=2[(5-3)\mathbf i+(1-(-4))\mathbf j]=(4\mathbf i+10\mathbf j)\,\text{N s}$ . Its magnitude is $\sqrt{4^2+10^2}=2\sqrt{29}\,\text{N s}$ .
2	The resultant impulse is $(-2\mathbf i-4\mathbf j)\,\text{N s}$ The change in velocity is $(-\mathbf i-2\mathbf j)\,\text{m s}^{-1}$ The final velocity is $(2\mathbf i-\mathbf j)\,\text{m s}^{-1}$ The final speed is $\sqrt5\,\text{m s}^{-1}$	4
(4 marks)		4
Notes
The two impulses combine to $(2\mathbf i-6\mathbf j)+(-4\mathbf i+2\mathbf j)=-2\mathbf i-4\mathbf j$ . Hence $\Delta\mathbf v=\mathbf J/m=-\mathbf i-2\mathbf j$ , so $\mathbf v=(3\mathbf i+\mathbf j)+(-\mathbf i-2\mathbf j)=2\mathbf i-\mathbf j$ . Its magnitude is $\sqrt{2^2+(-1)^2}=\sqrt5\,\text{m s}^{-1}$ .

Tier 2 · Standard

Mark scheme for FM1-1.2 Tier 2 · Standard
Question	Scheme	Marks
1	Speed $\dfrac{\sqrt5}{2}\,\text{m s}^{-1}$ Angle $\tan^{-1}2\approx63.4^\circ$	4
(4 marks)		4
Notes
$\mathbf v=\mathbf u+\mathbf J/m=(2\mathbf i-\mathbf j)+(-1.5\mathbf i+2\mathbf j)=(0.5\mathbf i+\mathbf j)\,\text{m s}^{-1}$ . Hence $\|\mathbf v\|=\sqrt{0.5^2+1^2}=\sqrt5/2$ , and $\tan\theta=1/0.5=2$ .
2	The final $\mathbf i$ component is $1+4/2=3\,\text{m s}^{-1}$ The speed condition gives the final $\mathbf j$ component as $4$ or $-4$ These alternatives give impulse $\mathbf j$ components $12\,\text{N s}$ and $-4\,\text{N s}$ respectively The positive impulse component selects $\mathbf v=(3\mathbf i+4\mathbf j)\,\text{m s}^{-1}$ $\mathbf J=(4\mathbf i+12\mathbf j)\,\text{N s}$ , in unit direction $(\mathbf i+3\mathbf j)/\sqrt{10}$	5
(5 marks)		5
Notes
The $\mathbf i$ component of impulse changes the $\mathbf i$ velocity by $4/2=2$ , so the final $\mathbf i$ component is $3$ . A speed of $5$ then gives final $\mathbf j$ component $4$ or $-4$ . The corresponding impulse $\mathbf j$ components are $2[4-(-2)]=12$ and $2[-4-(-2)]=-4\,\text{N s}$ . Only the first gives the impulse a positive $\mathbf j$ component. Thus $\mathbf v=3\mathbf i+4\mathbf j$ and $\mathbf J=4\mathbf i+12\mathbf j=4(\mathbf i+3\mathbf j)$ , whose unit direction is $(\mathbf i+3\mathbf j)/\sqrt{10}$ .
3	Using the $\mathbf i$ components, $-12=m(1-5)$ Hence $m=3\,\text{kg}$ The $\mathbf j$ components are consistent: $9=3[1-(-2)]$ The speed after the impulse is $\sqrt{1^2+1^2}=\sqrt2\,\text{m s}^{-1}$	4
(4 marks)		4
Notes
Using $\mathbf J=m(\mathbf v-\mathbf u)$ and comparing $\mathbf i$ components gives $-12=m(1-5)$ , so $m=3\,\text{kg}$ . The $\mathbf j$ equation then gives $9=3[1-(-2)]$ , confirming that the same mass satisfies both components. The final velocity is $\mathbf i+\mathbf j$ , so its speed is $\sqrt{1^2+1^2}=\sqrt2\,\text{m s}^{-1}$ .

Tier 3 · Hard

Mark scheme for FM1-1.2 Tier 3 · Hard
Question	Scheme	Marks
1	$\mathbf v=(\mathbf i+2\mathbf j)\,\text{m s}^{-1}$ with $\mathbf J=(-9\mathbf i+12\mathbf j)\,\text{N s}$ $\mathbf v=(-\mathbf i-2\mathbf j)\,\text{m s}^{-1}$ with $\mathbf J=-15\mathbf i\,\text{N s}$	5
(5 marks)		5
Notes
Write $\mathbf v=x\mathbf i+y\mathbf j$ . Perpendicularity gives $4x-2y=0$ , so $y=2x$ . Also $\mathbf J=3(\mathbf v-\mathbf u)=[3(x-4)\mathbf i+(6x+6)\mathbf j]$ . The magnitude condition gives $9(x-4)^2+(6x+6)^2=225$ , which simplifies to $45x^2+180=225$ . Thus $x=\pm1$ . Substitution gives the two stated velocity and impulse pairs.
2	For $P$ , $(2+a)^2+(1+b)^2=50$ , so $a^2+b^2+4a+2b=45$ For $Q$ , $(2+a/2)^2+(1+b/2)^2=85/4$ , so $a^2+b^2+8a+4b=65$ Subtracting the equations gives $2a+b=10$ Substitution gives $a^2+b^2=25$ and hence $(a-3)(a-5)=0$ The candidates are $(a,b)=(3,4)$ and $(5,0)$ The condition $b>0$ selects $\mathbf J=(3\mathbf i+4\mathbf j)\,\text{N s}$ Its magnitude is $5\,\text{N s}$ and its direction is $\tfrac35\mathbf i+\tfrac45\mathbf j$	7
(7 marks)		7
Notes
For $P$ , the speed condition is $(2+a)^2+(1+b)^2=50$ , so $a^2+b^2+4a+2b=45$ . For $Q$ , $(2+a/2)^2+(1+b/2)^2=85/4$ , so $a^2+b^2+8a+4b=65$ . Subtraction gives $2a+b=10$ ; substitution into the first equation gives $a^2+b^2=25$ . With $b=10-2a$ , this becomes $(a-3)(a-5)=0$ . The candidates are $(a,b)=(3,4)$ and $(5,0)$ , and $b>0$ selects $(3,4)$ . Its magnitude is $5$ and its unit direction is $(3\mathbf i+4\mathbf j)/5$ .
3	The impulse is $(-6t\mathbf i+8t\mathbf j)\,\text{N s}$ The final velocity is $[(4-3t)\mathbf i+(-1+4t)\mathbf j]\,\text{m s}^{-1}$ Parallelism gives $-1+4t=2(4-3t)$ Therefore $t=9/10\,\text{s}$ The final velocity is $(13\mathbf i/10+13\mathbf j/5)\,\text{m s}^{-1}$ The final speed is $13\sqrt5/10\,\text{m s}^{-1}$ The impulse is $(-27\mathbf i/5+36\mathbf j/5)\,\text{N s}$	7
(7 marks)		7
Notes
The impulse is $(-6t\mathbf i+8t\mathbf j)\,\text{N s}$ , so division by the mass gives final velocity $(4-3t)\mathbf i+(-1+4t)\mathbf j$ . Parallelism to $\mathbf i+2\mathbf j$ requires $-1+4t=2(4-3t)$ , hence $t=9/10$ . Substitution gives final velocity $13\mathbf i/10+13\mathbf j/5$ , whose $\mathbf i$ component is positive as required and whose magnitude is $13\sqrt5/10$ . The force-duration product gives the stated impulse.
4	$a/3=12/4$ because $\mathbf J_1$ is parallel to $3\mathbf i+4\mathbf j$ Hence $a=9$ Perpendicularity gives $(9)(-8)+12b=0$ Hence $b=6$ $\mathbf J_1=(9\mathbf i+12\mathbf j)\,\text{N s}$ and $\mathbf J_2=(-8\mathbf i+6\mathbf j)\,\text{N s}$ The final velocity is $\bigl(\tfrac94\mathbf i+\tfrac32\mathbf j\bigr)\,\text{m s}^{-1}$ The final speed is $3\sqrt{13}/4\,\text{m s}^{-1}$	7
(7 marks)		7
Notes
Parallelism to $3\mathbf i+4\mathbf j$ gives $a/3=12/4$ , so $a=9$ . The zero scalar product $\mathbf J_1\boldsymbol\cdot\mathbf J_2=-72+12b$ then gives $b=6$ . The initial momentum is $8\mathbf i-12\mathbf j$ , and adding the two impulses gives final momentum $9\mathbf i+6\mathbf j$ . Division by the mass gives the stated velocity, whose magnitude is $\sqrt{(9/4)^2+(3/2)^2}=3\sqrt{13}/4$ .