Edexcel M2 2024 October — Question 2

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2024
SessionOctober
PaperDownload PDF ↗
TopicMomentum and Collisions
TypeVector impulse: find velocity or speed after impulse
DifficultyStandard +0.3 This is a straightforward M2 impulse-momentum question requiring application of the impulse-momentum theorem in 2D and the kinetic energy formula. Students must set up two equations (one from KE change, one from impulse magnitude) and solve simultaneously, which is standard for this topic with no novel insight required.
Spec6.02d Mechanical energy: KE and PE concepts6.03a Linear momentum: p = mv6.03f Impulse-momentum: relation
  1. A particle \(Q\) of mass 3 kg is moving on a smooth horizontal surface.
Particle \(Q\) is moving with velocity \(5 \mathbf { i } \mathrm {~ms} ^ { - 1 }\) when it receives a horizontal impulse of magnitude \(3 \sqrt { 82 } \mathrm { Ns }\). Immediately after receiving the impulse, the velocity of \(Q\) is \(( x \mathbf { i } + y \mathbf { j } ) \mathrm { ms } ^ { - 1 }\), where \(x\) and \(y\) are positive constants. The kinetic energy gained by \(Q\) as a result of receiving the impulse is 138 J .
Find, in terms of \(\mathbf { i }\) and \(\mathbf { j }\), the velocity of \(Q\) immediately after receiving the impulse.
Question 2:
AnswerMarks Guidance
2Use of Impulse momentum equation M1
I = 3 ( x i + y j ) − 3  5 i
AnswerMarks
( = 3 ( x − 5 ) i + 3 y ) jA1
( )
AnswerMarks
I 2 = 9 ( x − 5 ) 2 + y 2 = 9  8 2 (=738)M1
Equation for change in KEM1
1 ( )
 3 x 2 + y 2 − 2 5 = 1 3 8
2
( )
AnswerMarks
x 2 + y 2 − 2 5 = 9 2A1
( x − 5 ) 2 + y 2 = 8 2
AnswerMarks
x 2 + y 2 = 1 1 7  1 0 x = 6 0DM1
v=6i+9j ( ms−1 )A1
(7)
AnswerMarks
ALT3 8 2 c o s   
Use of I =
3 8 2 s i n 
3 82cos x 5
I = =3  −3 
 
AnswerMarks
3 82sin y 0M1 A1
( )
AnswerMarks
I 2 =9 ( x−5 )2 + y2 =982M1
Equation for change in KE
1 ( )
 3 x 2 + y 2 − 2 5 = 1 3 8
AnswerMarks
2M1 A1
Leads to =83.66...
x=5+3 82cos=6
y=3 82sin=9
AnswerMarks
v=6i+9j ( ms−1 )DM1
A1
Notes
AnswerMarks
M1Find the difference in momenta. Dimensionally correct. Condone
subtraction in wrong order. Must use both components. (Ignore
3 82 if it appears on LHS)
AnswerMarks
A1Correct unsimplified expression for difference in momentum.
Condone subtraction the wrong way round.
AnswerMarks
M1Correct use of Pythagoras with 3 82 and both components of
impulse. Ignore poor i-j notation, eg i2 for this mark, if recovered by
correct subsequent working.
AnswerMarks
M1Use change in KE to produce an equation in terms of x and y.
Dimensionally correct, requires 2 KE terms of correct structure.
Condone subtraction in wrong order.
Ignore poor i-j notation eg (xi)2 + (yj)2 for this mark, if recovered
by correct subsequent working.
M0 For use of velocity.
AnswerMarks
A1Correct unsimplified equation
A0 for incorrect notation. A0 for subtraction the wrong way round.
AnswerMarks
DM1Dependent on all preceding M marks. Solve for x or y
A1Correct velocity only. Accept column vector. ISW if continue to
find speed.
AnswerMarks Guidance
QuestionScheme Marks 
Question 2:
2 | Use of Impulse momentum equation | M1
I = 3 ( x i + y j ) − 3  5 i
( = 3 ( x − 5 ) i + 3 y ) j | A1
( )
I 2 = 9 ( x − 5 ) 2 + y 2 = 9  8 2 (=738) | M1
Equation for change in KE | M1
1 ( )
 3 x 2 + y 2 − 2 5 = 1 3 8
2
( )
x 2 + y 2 − 2 5 = 9 2 | A1
( x − 5 ) 2 + y 2 = 8 2
x 2 + y 2 = 1 1 7  1 0 x = 6 0 | DM1
v=6i+9j ( ms−1 ) | A1
(7)
ALT | 3 8 2 c o s   
Use of I =
3 8 2 s i n 
3 82cos x 5
I = =3  −3 
 
3 82sin y 0 | M1 A1
( )
I 2 =9 ( x−5 )2 + y2 =982 | M1
Equation for change in KE
1 ( )
 3 x 2 + y 2 − 2 5 = 1 3 8
2 | M1 A1
Leads to =83.66...
x=5+3 82cos=6
y=3 82sin=9
v=6i+9j ( ms−1 ) | DM1
A1
Notes
M1 | Find the difference in momenta. Dimensionally correct. Condone
subtraction in wrong order. Must use both components. (Ignore
3 82 if it appears on LHS)
A1 | Correct unsimplified expression for difference in momentum.
Condone subtraction the wrong way round.
M1 | Correct use of Pythagoras with 3 82 and both components of
impulse. Ignore poor i-j notation, eg i2 for this mark, if recovered by
correct subsequent working.
M1 | Use change in KE to produce an equation in terms of x and y.
Dimensionally correct, requires 2 KE terms of correct structure.
Condone subtraction in wrong order.
Ignore poor i-j notation eg (xi)2 + (yj)2 for this mark, if recovered
by correct subsequent working.
M0 For use of velocity.
A1 | Correct unsimplified equation
A0 for incorrect notation. A0 for subtraction the wrong way round.
DM1 | Dependent on all preceding M marks. Solve for x or y
A1 | Correct velocity only. Accept column vector. ISW if continue to
find speed.
Question | Scheme | Marks
\begin{enumerate}
  \item A particle $Q$ of mass 3 kg is moving on a smooth horizontal surface.
\end{enumerate}

Particle $Q$ is moving with velocity $5 \mathbf { i } \mathrm {~ms} ^ { - 1 }$ when it receives a horizontal impulse of magnitude $3 \sqrt { 82 } \mathrm { Ns }$.

Immediately after receiving the impulse, the velocity of $Q$ is $( x \mathbf { i } + y \mathbf { j } ) \mathrm { ms } ^ { - 1 }$, where $x$ and $y$ are positive constants.

The kinetic energy gained by $Q$ as a result of receiving the impulse is 138 J .\\
Find, in terms of $\mathbf { i }$ and $\mathbf { j }$, the velocity of $Q$ immediately after receiving the impulse.

\hfill \mbox{\textit{Edexcel M2 2024 Q2}}
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