Question
P
A
Q
0.75 m
B
3.25 m
C i
The diagram shows two particles P and Q which lie on a line of greatest slope of a plane ABC. Particles
P and Q are each of mass m kg. The plane is inclined at an angle i to the horizontal, where sin i = 0.6 .
The length of AB is 0.75 m and the length of BC is 3.25 m. The section AB of the plane is smooth and the section BC is rough. The coefficient of friction between each particle and the section BC is 0.25 .
Particle P is released from rest at A. At the same instant, particle Q is released from rest at B.
(a) Verify that particle P reaches B 0.5 s after it is released, with speed 3 m s -1. [3]
(b) Find the time that it takes from the instant the two particles are released until they collide. [4]
The two particles coalesce when they collide. The coefficient of friction between the combined particle and the plane is still 0.25 .
(c) Find the time that it takes from the instant the particles collide until the combined particle reaches C. [5]
Additional page
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Mark scheme
7(a) ( m ) a = ( m ) g sin θ = ( m ) g × 0.6 [ a = 6] M1 For attempt at Newton’s second law; 2 terms; allow
sign errors; allow sin/cos mix, allow θ = 36.9 or
better.
Allow θ = 37. Allow for a = 6 seen/used.
v2 = 02 + 2 × 6 × 0.75 v = 3 A1 AG.
3 = [ 0] + 6t t = 0.5 A1 AG.
Allow A2 for use of a = 6 with any 2 of v = 3 ,
t = 0.5 and s = 0.75 to get the third value.
Alternative Method 1 for Question 7(a)
( m ) a = ( m ) g sin θ = ( m ) g × 0.6 [ a = 6] M1 For attempt at Newton’s second law; 2 terms; allow
sign errors; allow sin/cos mix, allow θ = 36.9 or
better.
Allow θ = 37. Allow for a = 6 seen/used.
0 + 6 × 0.5 = 3 A1 AG.
1 A1 AG.
0 × 0.5 + × 6 × 0.52 = 0.75
2 1
Allow A2 for 3 × 0.5 − × 6 × 0.52 = 0.75.
1 2
or 3 × 0.5 − × 6 × 0.52 = 0.75
2
Alternative Method 2 for Question 7(a)
1 M1 Attempt at conservation of energy, 2 terms,
( m ) v2 = ( m ) g × 0.75 × 0.6 dimensionally correct, allow sin/cos mix, allow
2
θ = 36.9 or better. Allow θ = 37.
v =3 A1 AG.
Use constant acceleration to get t = 0.5 A1 AG. Must show a = 6.
7(a) Alternative Method 3 for Question 7(a)
1 M1 1
( s = ) ( 0 + 3) × 0.5 Use s = ( u + v ) t with u = 0 with any 2 of v = 3,
2 2
1 t = 0.5 and s = 0.75.
or 0.75 = ( 0 + v ) × 0.5
2
1
or 0.75 = ( 0 + 3) t
2
Correctly identifies one value. A1
( s = ) 0.75 A1 AG.
or v = 3
or t = 0.5
7(a) Alternative Method 4 for Question 7(a)
( )
32 = 02 + 2a × 0.75 → a = 6 B1 Use constant acceleration to get an equation in a
only, using u = 0 and any 2 of v = 3 , t = 0.5 and
or 3 = ( 0 + ) 0.5a → a = 6 s = 0.75.
1 Or using v = 3 , t = 0.5 and s = 0.75.
or 0.75 = ( 0 × 0.5 + ) a × 0.52 → a = 6
2
1
or 0.75 = 3 × 0.5 − a × 0.52 → a = 6
2
1 1 M1 Use constant acceleration formula in an attempt to
3 = ( 0 + ) 6t , 0.75 = ( 0 × t + ) × 6 × t 2 , 0.75 = 3t − × 6 × t 2 find the third unused value using a = 6.
2 2
1 1 Or use constant acceleration formula in an attempt to
or s = ( 0 × 0.5 + ) × 6 × 0.52 , s = 3 × 0.5 − × 6 × 0.52 find u using a = 6.
2 2
( ) 1
or v = ( 0 + ) 0.5 × 6; v 2 = 02 + 2 × 6 × 0.75; 0.75 = 0.5v − × 6 × 0.52
2
1
or 3 = u + 6 × 0.5 , 32 = u 2 + 2 × 6 × 0.75 , 0.75 = 0.5u + × 6 × 0.52
2
Get correct unused value A1 Or u = 0 if using all 3 values to find a = 6.
AG.
3
7(b) For BC ( m ) a = ( m ) g sin θ − ( m ) g × 0.8 × 0.25 = ( m ) g × 0.6 − ( m ) g × 0.8 × 0.25 [ a = 4] *M1 For attempt at Newton’s second law; 3 terms; allow
sign errors; allow g missing; allow sin/cos mix, allow
θ = 36.9 or better.
1 1 A1 For either s P or sQ .
For P , sP = 3t + × 4t 2 For Q , sQ = × 4 ( t + 0.5)
2
2 2 t is the time after P arrives at B.
1 1
Or for P , sP = 3(T − 0.5) + × 4 (T − 0.5) For Q , sQ = × 4T 2 T is the time from when both particles released.
2
2 2
1 1 DM1 For use of sP = sQ ; s P and sQ of correct form.
3t + × 4t 2 = × 4 ( t + 0.5) 3t + 2t 2 = 2t 2 + 2t + 0.5
2
2 2 Using their a ≠ their 6 from part (a), a ≠ ± g.
1 1
Or 3(T − 0.5) + × 4 (T − 0.5) = × 4T 2
2
Using same a in both s P and sQ .
2 2
( )
3T − 1.5 + 2 T 2 − T + 0.25 = 2T 2
t = 0.5, so total time = 1 s or time T = 1 s A1
7(b) Alternative Method for Question 7(b)
For BC ( m ) a = ( m ) g sin θ − ( m ) g × 0.8 × 0.25 = ( m ) g × 0.6 − ( m ) g × 0.8 × 0.25 [ a = 4] *M1 For attempt at Newton’s second law; 3 terms; allow
sign errors; allow g missing; allow sin/cos mix, allow
θ = 36.9 or better.
1 A1 For either. t is the time after P arrives at B.
For P sP = 3t + × 4t 2
2
1
Or Q has speed 2 m s–1 after 0.5 s, so sQ = 2t + × 4t 2
2
1 1 1 DM1 1
3t + × 4t 2 = 2t + × 4t 2 + × 4 × 0.52 For use of sP = sQ ± their × 4 × 0.52 ; s P or sQ of
2 2 2 2
correct form.
Using their a ≠ their 6 from part (a), a ≠ ± g.
Using same a in both s P or sQ .
t = 0.5 so total time = 1 s A1
Alternative Method 2 for Question 7(b): Using relative velocity
For BC ( m ) a = ( m ) g sin θ − ( m ) g × 0.8 × 0.25 = ( m ) g × 0.6 − ( m ) g × 0.8 × 0.25 [ a = 4] *M1 For attempt at Newton’s second law; 3 terms; allow
g missing; allow sign errors; allow sin/cos mix.
[In 0.5 s] Q has speed 2 m s–1 and has moved 0.5 m A1 For both.
0.5 DM1 Attempt at time from relative velocity using their 0.5
t= m and their 2 m s–1 .
3− 2
t = 0.5 so total time = 1 s A1
4
7(c) 1 B1 For either.
Immediately before the collision speed of P = 3 + 4 × = 5 ms −1
2
Speed of Q [ = 0 + 4 × 1] = 4 m s–1
5( m) + 4 ( m) = ( m) + ( m) × v [ v = 4.5] *M1 Use of conservation of momentum; 4 non-zero
terms; allow sign errors; allow their 4m s–1 ( ≠ 0 or 2 )
and 5m s–1 ( ≠ 3 ) .
Use of mg then withhold final A mark.
1 B1 May be implied by 1.25 m.
Distance from B at collision = × 4 × 12 = 2 m
2
1
OR = 3 × 0.5 + × 4 × 0.52 = 2 m
2
1 DM1 Use of constant acceleration with u = their 4.5,
1.25 = 4.5t + × 4t 2
2 a = their 4 ( ≠ ± g ) and s = 3.25 − their 2,
s ≠ 3.25, s ≠ 0.75, s ≠ 4, s ≠ 2.75, s ≠ 0.5.
If using two formulae, must be a complete method to
get an equation in t only.
t = 0.25s only A1 A0 if from use of mg in conservation of momentum.
5