Question 7:
--- 7(a) ---
7(a) | CLM | M1
1 0 m u − 6 m u = 5 m x + 2 m y
( 4 u = 5 x + 2 y ) | A1
Impact law | M1
y − x = 5 u e | A1
4 u = 5 x + 2 x + 1 0 u e ( = 7 x + 1 0 u e ) | DM1
( x  0  ) 1 0 u e  4 u | DM1
2
 0 „ e  o.e.
5 | A1
(7)
7(a)
M1 | Form CLM equation. All terms required. Mass and velocity correctly
paired. Dimensionally consistent, condone consistent additional g in
each term. Condone sign errors.
A1 | Correct equation or equivalent. Condone consistent additional g in
each term. Condone  y .
M1 | Use Impact Law. Dimensionally correct. Used the right way round
(separation and approach must not be interchanged). Condone sign
errors.
A1 | Correct equation or equivalent. Directions of P and Q after impact
must be consistent with CLM.
DM1 | Dependent on the preceding M marks. Eliminate velocity of Q to
form an equation in x, e and u only.
 4 u − 1 0 e u 
x =
7
DM1 | Dependent on all preceding M marks. May be implied by e  0 . 4
Use direction of P to form an inequality in e (and u). Use correct
inequality for their diagram: if they had P changing direction should
now be using x0
A1 | Both ends required 0 „ e  0 .4 but condone 0  e  0 .4
--- 7(b) ---
7(b) | Impulse momentum equation | M1
6 0
m u = 2 m ( 3 u − ( − y ) )
7
or
6 0
− m u = 5 m ( x − 2 u )
7 | A1
Solve to find a correct expression for either x or y
 9 2 
y = u , x = u b u t m a y n o t b e s e e n e x p l i c i t l y
7 7 | DM1
Use both impulse equations
and impact law with their x
and y to form an equation in
e (and u).
9 2
5 u e = u − u
7 7 | Using CLM and impact law from part
(a) to form an equation in e (and u).
For example
2 u 4 u − 1 0 e u
=
7 7 | M1
1  9 2  1
e = u − u = *
5 u 7 7 5 | A1*
(5)
7(b)
M1 | Impulse-momentum equation. Dimensionally correct, using the
correct mass and velocity pair for a single particle. Must be
subtracting momenta but allow incorrect order. May use working in
4 u − 1 0 e u 2 5 e u + 4 u
terms of e from (a) x = y =
7 7
A1 | At least one correct unsimplified equation.
DM1 | 9
Solve impulse-momentum equation to find either y = u or
7
2
x = u
7
M1 | Complete method to form an equation in e (and u) only with the usual
rules for CLM and Impact Law.
A1* | Obtain given answer from complete and correct working.
A0 for 0.2
--- 7(c) ---
7(c) | 2 3
7 u 7 u
P5 Q2
m m
x
y
1 9  3  Speed of Q =  u  = u 7 7
3 7  7  | B1ft
1 60
Magnitude of impulse =  mu
7 7 | M1
60
= mu
49 | A1
(3)
(15)
Notes
7(c)
B1 | 1 1
Follow their y from part (b). If y is given in terms of e then e=
3 5
must be substituted at some point. Seen or implied. May appear on
diagram. Accept .
M1 | Complete method using (a) and (b) and working from first principles
to find the impulse in the second collision between P and Q.
Must see:
• Usual rules for CLM and Impact Law to find V or V after
p q
second collision.
CLM
1 0 m u 6 m u
− = 5 m V + 2 m V
7 7 p q
Impact Law
1  2 u 3 u 
 + = V − V
5 7 7 q p
• Then use of Impulse – momentum equation with usual rules to
find magnitude of impulse.
Impulse – momentum
 9 u 3 u   2 u 2 u 
Either I = 2 m − − or I = 5 m −
4 9 7 4 9 7
A1 | 1.2mu or better (1.2244897…)
PMT
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