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# PMT

GCE Examinations

Mechanics
Module M1

Paper K
MARKING GUIDE

## This guide is intended to be as helpful as possible to teachers by providing

concise solutions and indicating how marks should be awarded. There are
obviously alternative methods that would also gain full marks.

Method marks (M) are awarded for knowing and using a method.

Accuracy marks (A) can only be awarded when a correct method has been used.

## Written by Shaun Armstrong & Chris Huffer

 Solomon Press
These sheets may be copied for use solely by the purchaser’s institute.
PMT

## 1. (a) impulse = ∆ mom = 800(0 – 15) = – 12 000 ∴ mag. = 12 000 Ns M1 A1

12000
(b) Ft = 12 000, so t = 60000 = 0.2 s M1 A1

## (c) use v = u + at with v = 0, u = 15 t = 0.2 M1

0 = 15 + 0.2a ∴ a = – 75 so decel. = 75 m s-2 M1 A1 (7)

## 2. (a) moments about O (clockwise +ve): 90(1.2) + 70(1.2) – 80(1.2) – 60(1.2) M1

= 20(1.2) = 24 Ns (clockwise) M1 A1

## (b) moments about O : 90(d) + 70(1.2) – 80(1.2) – 60(1) = 0 M2 A1

90d = 72 ∴ d = 0.8 m ∴ x = 0.4 m M1 A1 (8)

3. (a) r = 4ti m A1
s = (30i – 60j) + (– 8ti + 24tj) M1
(30 – 8t)i + (24t – 60)j m A1

## (b) they will collide if coeffs. of i and j in r and s are equal B1

4t = 30 – 8t and 24t – 60 = 0 M1
both are satisfied when t = 52 so ball hits batsman M1 A1
batsman is at (4 × 2.5)i = 10i A1

## (c) ball travelling fast ∴ air resistance significant B1

ball will be affected by gravity ∴ not horizontal (may go over batsman) B1 (10)

## 4. (a) cons. of mom. 4m × 2 – 3m × 2 = 0 + 3mv M1

2m = 3mv so v = 23 m s-1 M1 A1

(b) R = mg; – F = ma M1
− µR − µmg
but F = µR; so a = m = m = – µg M1 A1
use with u = 23 , v = 0, s = 0.2 M1
v = u + 2as, ∴ 0 =
2 2 4
9
– 0.4µg M1
µ= 10
9g
= 0.113 (3dp) M1 A1 (10)

5. (a) R 50

20
5° 70g

## resolve // to plane: 50 – 20 – 70gsin5° = 70a M1 A1

a = – 0.43, ∴ decel. = 0.43 m s-2 (2dp) M1 A1

(b) use of s = ut + 1
2 at2 with u = 3 for t = 4 and t = 5 M1
to give s = 12 + 8a and s = 15 + 12.5a M1 A1
∴ 15 + 12.5a = 12 + 8a + 12 M1
4.5a = 9 ∴ a = 2 M1 A1
use v = u + at with u = 3, a = 2, t = 5 M1
v = 3 + 2 × 5 = 13 m s-1 A1 (12)

 Solomon Press
M1K MARKS page 2
PMT

## 6. (a) eqn. of motion for A: T = 5a (1) M1

eqn. of motion for Q: 2g – T = 2a (2) M1
2g
(1) + (2) gives 2g = 7a i.e. a = 7
A1
10
from (1), T = 5a = 7
g N A1
so force on pulley = T 2 + T 2 = T 2 M1
10 2
= 7
gN M1 A1

(b) s = 0.3, u = 0, a = 2
7
g use v2 = u2 + 2as M1
6 6
v2 = 35 g i.e. v = 35
g = 1.68 = 1.30 m s-1 (3sf) M1 A1

## (c) B has 0.2 m left to fall B1

6
for B: u2 = 35 g, s = 0.2, a = g use v2 = u2 + 2as M1
v2 = 6
35
g + 2g(0.2) ∴ v2 = 5.6; v = 2.4 m s-1 (1dp) M1 A1 (14)

7. R T

25g
35°

## (a) resolve // to plane: Tcos15 – 25gsin35 = 0 M2

so T = 145.48 = 145 N (3sf) A1

## (b) resolve perp. to plane: R + Tsin15 – 25gcos35 = 0 M1 A1

R = 25gcos35 – 145.48 × sin15 = 163.038 = 163 N (3sf) M1 A1
(c) resolve // to plane: 200cos15 – F – 25gsin35 = 0 M2
F = 52.7 N (3sf) down the slope A2

(d) decrease B1
e.g. perp. to plane, same force “down”, more from T “up” ∴ R less B2 (14)

Total (75)

 Solomon Press
M1K MARKS page 3
PMT

## Question no. 1 2 3 4 5 6 7 Total

Topic(s) impulse, moments rel. posn. cons. of uniform connected statics
uniform i, j mom. accel., bodies
accel friction F = ma

Marks 7 8 10 10 12 14 14 75
Student

 Solomon Press
M1K MARKS page 4
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