Challenging +1.2 This is a standard two-stage mechanics problem combining energy conservation with collision theory. While it requires careful bookkeeping through multiple steps (finding P's speed before collision, applying conservation of momentum and restitution formula, then using energy conservation for Q's motion), each individual technique is routine for Further Mechanics students. The 'just reaches the rim' condition provides a clear constraint that makes the problem straightforward to set up and solve algebraically.
\includegraphics{figure_6}
The rim of a smooth hemispherical bowl is a circle of centre O and radius \(a\). The bowl is fixed with its rim horizontal and uppermost. A particle P of mass \(m\) is released from rest at a point A on the rim as shown in Fig. 6.
When P reaches the lowest point of the bowl it collides directly with a stationary particle Q of mass \(\frac{1}{2}m\). After the collision Q just reaches the rim of the bowl.
Find the coefficient of restitution between P and Q. [7]
NOTE THAT THERE ARE TWO PAGES FOR THIS QUESTION – PLEASE ENSURE BOTH ARE CHECKED
Question 6:
6 | 1
mga mv2
2
v2 2ga or v 2ga
1
mv mv m 2ga
Q P
2
v v e 0 2ga
Q P
11 1
v Q 2gaor 2 2 m v Q 2 2 mga
1 2ga
v 2ga or v 2e
P 2 P 3
1
2ga 2ga e 2ga or
2
2
11 2 1
m 2ga1e mga
22 3 2
1
e
2 | M1*
A1
M1*
M1*
B1
M1dep*
A1
[7] | 3.3
1.1
3.3
3.3
3.3
2.1
2.2a | Equating KE gained for P with the PE
lost
v orv2 of P before impact
Attempt at conservation of momentum
– correct number of terms – must be
using their speed of P before impact in
terms of g and a
Attempt at Newton’s experimental law
– correct number of terms and
consistent with CLM – must be using
their speed of P before impact in terms
of g and a
www where v is the speed of Q after
Q
impact – for reference
v 2 2ga1e
Q 3
Where first v expression comes from
P
1
mv mv m 2gaand v 2ga
2 Q P Q
Setting up an equation involving g, a
and e – dependent on all previous M
marks only and must have found v
P
and/or v from a correct method
Q | v is the speed of P
before impact
Correct masses for P
and Q – note that v
Q
may already have been
substituted
This mark may be
earned later after v and
Q
v have been found
P
1
Condone mv 2 mga
2 Q
No marks for these
Either substituting into
NEL or considering
energy of Q after
collision
NOTE THAT THERE ARE TWO PAGES FOR THIS QUESTION – PLEASE ENSURE BOTH ARE CHECKED
\includegraphics{figure_6}
The rim of a smooth hemispherical bowl is a circle of centre O and radius $a$. The bowl is fixed with its rim horizontal and uppermost. A particle P of mass $m$ is released from rest at a point A on the rim as shown in Fig. 6.
When P reaches the lowest point of the bowl it collides directly with a stationary particle Q of mass $\frac{1}{2}m$. After the collision Q just reaches the rim of the bowl.
Find the coefficient of restitution between P and Q. [7]
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2019 Q6 [7]}}