CAIE A-Level Mathematics AS 4.3.2 Collisions and Momentum

[Maximum number: 11]

Three particles A, B and C of masses $5 \mathrm{~kg}, 1 \mathrm{~kg}$ and 2 kg respectively lie at rest in that order on a straight smooth horizontal track X Y Z. Initially A is at X, B is at Y and C is at Z. Particle A is projected towards B with a speed of $6 \mathrm{~ms}^{-1}$ and at the same instant C is projected towards B with a speed of $v \mathrm{~ms}^{-1}$ . In the subsequent motion, A collides and coalesces with B to form particle D. Particle D then collides and coalesces with C to form particle E and E moves towards Z.

(a)

Show that after the second collision the speed of E is $\frac{15-v}{4} \mathrm{~ms}^{-1}$ .

[ 3 ]

Attempt at conservation of momentum for the 1st collision
$\left[5 \times 6=(5+1) v_{D}\right]$

*M1

3 non-zero terms;
allow sign errors;
using correct masses. For reference $v_{D}=5$ .

If m g v used, allow M1 M1 A0 max.

Attempt at conservation of momentum for the 2nd collision
$\left[(5+1)\left(\text { their } v_{D}\right)-2 v=(5+1+2) v_{E}\right]$
6 non-zero terms;
allow sign errors;
using correct masses;
allow their numerical $v_{D}$ .

Marking guidance:

Allow $v=v_{E}$ for this mark.

Note: $5 \times 6-2 v=(5+1+2) v_{E}$ is M2 .

If m g v used, allow M1 M1 A0 max.

$\left(v_{E}=\right) \frac{15-v}{4}$
AG

Must in terms of v, as v is given in the question or explicitly defined their letter used as v.

Do not allow $v=v_{E}$ for this mark.

Any error seen is A0 but condone saying 'divide by 2 ' or equivalent.

If m g v used, allow M1 M1 A0 max.

(b)

The total loss of kinetic energy of the system due to the two collisions is 63 J .

Use the result from (a) to show that v=3.

[ 3 ]

The initial kinetic energy is $90+v^{2}$ , and the final kinetic energy is
$\frac12(5+1+2)\left(\frac{15-v}{4}\right)^{2}.$
Equating the loss to 63 J gives $3v^{2}+30v-117=0$ . The physically valid solution is v=3.

(c)

It is given that the distance X Y is 36 m and the distance Y Z is 98 m .

[ 5 ]

(i)

Find the time between the two collisions.

[ 4 ]

Time A to $B=6 \mathrm{~s}$
Distance $B C=98-3 \times($ their 6)[=80]

*B1FT

FT their 6 which MUST come from 6 t=36.

Use sum of distance moved by D and distance moved by C is 80 m
$[(\text { their } 5) t+3 t=\text { their } 80]$

OR use distance moved by C divided by relative velocity
$\left[\frac{80}{(\text { their } 5)+3}\right]$
Using their $v_{D}$ from part (a).
$v_{D} \neq 6$ or 3 and their $80 \neq 98$ .

Time $=10 \mathrm{~s}$
Do not ISW.

(ii)

Find the time between the instant that A is projected from X and the instant that E reaches Z.

[ 1 ]

$\left[6+10+\frac{3 \times 10+3 \times 6}{3}=\right] 32 \mathrm{~s}$

CAIE A-Level Mathematics AS 4.3.2 Momentum Questions

Exam points

CAIE A-Level Mathematics AS 4.3.2 Momentum Questions question 1

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)