CAIE A-Level Physics 3.3 Linear Momentum & Its Conservation Question Bank

Practise conserving vector momentum in collisions, explosions and decays and distinguishing elastic from inelastic outcomes.

Covered syllabus points

3.3.1—The principle of conservation of momentum 3.3.2—The principle of conservation of momentum to solve simple problems 3.3.3—That, for an elastic collision, total kinetic energy is conserved and the 3.3.4—That, while momentum of a system is always conserved in interactions between

[Maximum number: 10]

(a)

State the principle of conservation of momentum.

[ 2 ]

(b)

Ball A moves with speed v along a horizontal frictionless surface towards a stationary ball B, as shown in Fig. 3.1.

Fig. 3.1

Ball A has mass 4.0 kg and ball B has mass 12 kg .
The balls collide and then move apart as shown in Fig. 3.2.
Ball A has velocity $6.0 \mathrm{~ms}^{-1}$ at an angle of $\theta$ to the direction of its initial path.
Ball B has velocity $3.5 \mathrm{~ms}^{-1}$ at an angle of $30^{\circ}$ to the direction of the initial path of ball A.

[ 8 ]

(i)

By considering the components of momentum at right-angles to the direction of the initial path of ball A, calculate $\theta$ .

\theta=

[ 3 ]

p=m v

C1

$(4.0 \times 6.0 \times \sin \theta)-\left(12 \times 3.5 \times \sin 30^{\circ}\right)=0$

or
$\left(m_{A} v_{A} \times \sin \theta\right)-\left(m_{B} v_{B} \times \sin 30^{\circ}\right)=0$

M1

$\theta=61^{\circ}$

A1

(ii)

Use your answer in (i) to show that the initial speed v of ball A is $12 \mathrm{~ms}^{-1}$ . Explain your working.

[ 2 ]

shows the horizontal momentum component of ball A or of ball B as ( $4.0 \times 6.0 \times \cos \theta$ ) or ( $12 \times 3.5 \times \cos 30^{\circ}$ )

C1

$\left(4.0 \times 6.0 \times \cos 61^{\circ}\right)+\left(12 \times 3.5 \times \cos 30^{\circ}\right)=4.0 \mathrm{v}$ so $\mathrm{v}=12\left(\mathrm{~ms}^{-1}\right)$

A1

(iii)

By calculation of kinetic energies, state and explain whether the collision is elastic or inelastic.

[ 3 ]

initial $E_{\mathrm{K}}\left(=1 / 2 \times 4.0 \times 12^{2}\right)=290(288)(\mathrm{J})$

M1

final $E_{\mathrm{K}}\left(=1 / 2 \times 4.0 \times 6.0^{2}+1 / 2 \times 12 \times 3.5^{2}\right)=150(145.5)(\mathrm{J})$

M1

(initial $E_{\mathrm{K}}>$ final $E_{\mathrm{K}}$ ) so inelastic [both M1 marks required to award this mark]

A1

[Maximum number: 6]

(a)

Two blocks travel directly towards each other along a horizontal, frictionless surface. The blocks collide, as illustrated in Fig. 3.1.

Fig. 3.1

Block A has mass 3 M and block B has mass M.
Before the collision, block A moves to the right with speed $0.40 \mathrm{~ms}^{-1}$ and block B moves to the left with speed $0.25 \mathrm{~ms}^{-1}$ .
After the collision, block A moves to the right with speed $0.20 \mathrm{~m} \mathrm{~s}^{-1}$ and block B moves to the right with speed v.

[ 6 ]

(i)

Determine speed v.

[ 3 ]

p=m v or $3 M \times 0.40$ or $M \times 0.25$ or $3 M \times 0.2$ or M v

C1

$(3 M \times 0.40)-(M \times 0.25)=(3 M \times 0.2)+M v$

C1

$\begin{aligned} v =(3 \times 0.40)-0.25-(3 \times 0.2) =0.35 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned}$

A1

(ii)

Calculate, for the blocks,
1. the relative speed of approach,
relative speed of approach = $\mathrm{m s}^{-1}$
2. the relative speed of separation.
relative speed of separation = $\mathrm{m s}^{-1}$

[ 2 ]

1. relative speed of approach =0.40+0.25
$=0.65 \mathrm{~ms}^{-1}$

A1

2. relative speed of separation =0.35-0.20
$=0.15 \mathrm{~ms}^{-1}$

A1

(iii)

Use your answers in (b)(iii) to state and explain whether the collision is elastic or inelastic.

[ 1 ]

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Covered syllabus points

3.3 Linear momentum and its conservation question 1

Question (a)

Question (b)

Question (i)

Question (ii)

Question (iii)

3.3 Linear momentum and its conservation question 2

Question (a)

Question (i)

Question (ii)

Question (iii)