Exam-style paper B
1. By conservation of momentum
5. i) For B: 0.9g – T = 0.9a
m × 2 + 3 × (–3) = m × (–0.5) + 3 × 0.5
m = 4.2
Hence 0.6g = 1.5a so a = 4 m s−2 and T = 5.4 N
2. Resolving horizontally:
5 + 12 cos 45° – 10 cos 60° – 4 cos 75° = 7.45 N
Resolving vertically:
10 sin 60° – 12 sin 45° – 4 sin 75° = −3.69 N
Using Pythagoras’ theorem:
3.69
in the direction tan−1
= 26.3° below horizontal
7.45
to the right.
3. If cos θ = 0.96, sin θ = 0.28
ii) F + 0.45 cos θ = 0.35g sin θ ⇒ F = 0.548 N
0.548
= 0.169
3.23
First segment from (0, 0) to (T1, V ) with slope
= 0.4; second segment from (T1, V ) to (T1 + T2, V );
third segment from (T1 + T2, V ) to (655, 0).
V
V
5V
=
and final deceleration =
40
0.4
2
(T2 + 655 )
iii) Area of trapezium: V
2
5V
T2 = 655 – 40 – T1 = 615 –
2

V
V
+ 655
5000 =  615 −


2
0.4
66 000
= 5500 N
12
1
5500 – 12k – 5000g ×
= 0 so 12k = 3000
20
6. i) Power = F × v so F =
1
66000
+ 5000g ×
– 250v = 0
20
v
ii)
5V 

2 
4000 = 508V – V 2 ⇒ V 2 – 508V + 4000 = 0
V = 8 or 500
66000
+ 2500 – 250v = 0
v
66000 + 2500v – 250v 2 = 0
v 2 – 10v – 264 = 0, since v > 0, v = 22 m s−1
7. i) 100 cos 20° × 30 = 2819.1 J
ii) Increase in PE from 0: 25 × g × 30 sin 5° = 653.67 J
ii)2819.1 – 653.67 – (work done against resistance) =
2165.4 – (30 × 70) = 65.410
= 5000
4000 = V(508 – V )
−0.6g sin 30° = 0.6a' ⇒ a' = −5 m s−2
4. i)Graph made up of three straight-line segments:

iii) T = 0 so acceleration of A is now:
therefore k = 250
i) R = 0.35g cos θ – 0.45 sin θ = 3.23 N
10 000 = V  1270 −
v = 2 × 4 × 0.5 = 2 m s−1
⇒ s = 0.4 m
= 7.452 + 3.69 2 = 8.31 N
ii) v = at, so T1 =
ii) Using v 2 = u2 + 2as on B:
Using v 2 = u2 + 2as on A: 0 = 4 – 10 s
Resultant force
iii) μ =
For A: T – 0.6g sin 30° = 0.6a
Increase in KE = 65.4 N
therefore
v=
65.4
= 2.29 m s−1
12.5
Alternatively
100 cos 20° – 70 – 25g sin 5° = 25a so
a = 0.0872… m s−2
Using v 2 = u2 + 2as: v = 2 × 0.087... × 30
If V = 500 then T1 = 1250
= 2.29 m s−1
But T1 < 655
So V = 8 m s–1
© Oxford University Press 2018: this may be reproduced for class use solely for the purchaser’s institute
Exam-style paper B
1