Challenging +1.2 This is a multi-stage mechanics problem requiring: (1) finding when/where particles meet using SUVAT equations, (2) applying conservation of momentum for the collision, (3) using the coefficient of restitution formula, and (4) calculating subsequent motion to ground. While it involves several standard techniques and careful bookkeeping across multiple stages, each individual step uses routine Further Maths mechanics methods without requiring novel insight. The 10-mark allocation reflects the length rather than conceptual difficulty.
A particle P of mass 2 kg is projected vertically upwards from horizontal ground with an initial speed of \(14 \text{ m s}^{-1}\). At the same instant a particle Q of mass 8 kg is released from rest 5 m vertically above P. During the subsequent motion P and Q collide. The coefficient of restitution between P and Q is \(\frac{11}{14}\).
Determine the time between this collision and P subsequently hitting the ground. [10]
Collision occurs at a distance 4.375 above the ground
Before collision P’s velocity: ±(14−g× 5) (=±10.5)
14
Before collision Q’s velocity: ±(g× 5) (=±3.5)
14
2v +8v =2(10.5)+8(−3.5)
P Q
v −v =−11(10.5−(−3.5))
P Q 14
v =−9.5,(v =1.5)
P Q
4.375=9.5t+0.5gt2
Answer
Marks
t = 0.384 (s)
B1
B1
M1*
M1*
M1dep*
A1FT
A1FT
A1
M1
A1
Answer
Marks
[10]
3.1b
3.1b
3.3
1.1
3.3
1.1
1.1
1.1
3.4
Answer
Marks
2.2a
awrt 0.357 or allow 0.36 www
This could appear at any stage of their working – or for
0.625 (distance that Q travels before collision)
Use of v=u+at with u = 14 and a=±g and their t
which must be less than 0.419 – if using v2 =u2 +2as
their s < 5
Using their t < 0.419 or from v2 =u2 +2aswith u = 0,
a=+g and their s < 5
Use of either CLM or NEL for the collision between P
and Q (correct form (so e.g. e must be with the approach
speed) and correct number of terms but allow sign errors)
With their 10.5 and 3.5 (but must be different signs)
Consistent with CLM with their 10.5 and 3.5 – so must be
considering ±(10.5+3.5)for their 10.5 and 3.5
Only v required – allow positive value (9.5) www
P
Using s=ut+0.5at2 (oe) using correct 4.375 and their u
to set up a 3TQ in t – dependent on all previous M
marks. Allow wrong sign(s) for M mark
BC negative root need not be rejected but 0.384 must be
their final answer – awrt 0.384 or allow 0.38 www
Question 10:
10 | Time to collision is 5 (= 0.35714…)
14
Collision occurs at a distance 4.375 above the ground
Before collision P’s velocity: ±(14−g× 5) (=±10.5)
14
Before collision Q’s velocity: ±(g× 5) (=±3.5)
14
2v +8v =2(10.5)+8(−3.5)
P Q
v −v =−11(10.5−(−3.5))
P Q 14
v =−9.5,(v =1.5)
P Q
4.375=9.5t+0.5gt2
t = 0.384 (s) | B1
B1
M1*
M1*
M1dep*
A1FT
A1FT
A1
M1
A1
[10] | 3.1b
3.1b
3.3
1.1
3.3
1.1
1.1
1.1
3.4
2.2a | awrt 0.357 or allow 0.36 www
This could appear at any stage of their working – or for
0.625 (distance that Q travels before collision)
Use of v=u+at with u = 14 and a=±g and their t
which must be less than 0.419 – if using v2 =u2 +2as
their s < 5
Using their t < 0.419 or from v2 =u2 +2aswith u = 0,
a=+g and their s < 5
Use of either CLM or NEL for the collision between P
and Q (correct form (so e.g. e must be with the approach
speed) and correct number of terms but allow sign errors)
With their 10.5 and 3.5 (but must be different signs)
Consistent with CLM with their 10.5 and 3.5 – so must be
considering ±(10.5+3.5)for their 10.5 and 3.5
Only v required – allow positive value (9.5) www
P
Using s=ut+0.5at2 (oe) using correct 4.375 and their u
to set up a 3TQ in t – dependent on all previous M
marks. Allow wrong sign(s) for M mark
BC negative root need not be rejected but 0.384 must be
their final answer – awrt 0.384 or allow 0.38 www
A particle P of mass 2 kg is projected vertically upwards from horizontal ground with an initial speed of $14 \text{ m s}^{-1}$. At the same instant a particle Q of mass 8 kg is released from rest 5 m vertically above P. During the subsequent motion P and Q collide. The coefficient of restitution between P and Q is $\frac{11}{14}$.
Determine the time between this collision and P subsequently hitting the ground. [10]
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2024 Q10 [10]}}