Chapter 4
Temperature outside.
4.1Cost.
Heat saving
4.2See chapter 3. Height of thrower.
4.3Angle of projection. Speed of throw.
4.4Number of lifts available.
Correct lift positions, number of other users, speed of operation of lifts.
(a)(i) No. (ii) Yes. (iii) No. (iv) Yes.
(b)(i) Yes. (ii) No. (iii) No. (iv) No.
(c)(i) No. (ii) No. (iii) Yes. (iv) No.
(d)(i) No. (ii) No. (iii) Yes. (iv) No.
4.5
(e)(i) No. (ii) No. (iii) Yes. (iv) No.
(f)(i) Yes. (ii) No. (iii) No. (iv) No.
(g)(i) No. (ii) Yes. (iii) No. (iv) Yes.
(h)(i) No. (ii) Yes. (iii) No. (iv) No.
(a)(i) Yes. (ii) No. (iii) Yes.
4.6(b) (i) No. (ii) Yes. (iii) No.
(c)(i) Yes. (ii) Yes. (iii) Yes.
(i) a gives a vertical shift.
b affects magnitude of expression.
c affects rate of decay.
4.7
(ii) a magnitudes the expression. b shifts maximum point.
c gives a vertical shift.
4.8(c) is correct.
4.9(e) is correct.
4.10kLU 2 / D.
[ k ] = ML −1 . kg m −1 .
4.11A + B sin(π t /12) + c sin(π t /4380).
4.12A ( x ) = a/x, B ( y ) = b/y where a/x + b/y = c.
4.13A sin 2 (π t /6) exp(−0.1 t ).
(i)c is the smallest term, and 
the largest term.
4.14(ii) ab is the smallest term, and 
the largest term.
(iii)c is the smallest term, and 
the largest term.
4.15(i) x 2 + a. (ii) 
(iii) 
4.16(i) 0.01 x 3 . (ii) 
(iii) 
4.17
(i) 0.001 + 0.0001 x. (ii) x. (iii) 4.18 1.79 m s −1
4.19 1 N = 10 5 dyn = 7.23 pdl.
4.20 1 million ft 3 day −1 = 0.3278 m 3 s −1 .
4.21 The track with a perimeter of 440 yd is longer by 2.34 m. 4.22 7.27 × 10 −5 rad s −1 , 6.67 × 10 4 mph.
4.23 Nm 2 kg −2
4.24 ML −7 T −2
4.25(a) Correct (b) Error (c) Correct (d) Error
4.26(a) Error (b) Correct (c) Error
4.27Incorrect
4.28
and c = 0. Thus u = 
Chapter 5
5.2F n +1 = 1.15 F n − 2000 − kP n , P n +1 = 0.99997 P n
5.3A, E, G
5.4E, G
5.5(a) X n +1 = (1 + r /100) X n − 12 m
5.6A n +1 = A n + 15, B n +1 = B n − 9, P n +1 = P n + 5 − 790 P n / A n , Q n +1 = Q n +1 + 790 P n / A n − 800 Q n / B n
(a)X n +1 = (19 X n + 100)/21
5.7(b) X n = 50[1 − (19/21) n ], Y n = 100 − X n
(c)X n and Y n become 50
Chapter 6
6.1 
t measured in years, N in thousands.
6.2T ( t ) = 18 + 42 exp(−0.09060 t ). Cools to 30°C in 13.82 min. T (10) = 34.97°C.
6.3V ( t ) = (1.279 − 10.0399 t ) 3 . V is in m 3 , t is in h.
6.4Y = α 2 /4β.
(a)False. (b) False. (c) False. (d) False.
6.5(e) (i) No. (ii) Velocity zero. (iii) No.
(f)(i) No. (ii) Yes. (iii) Yes. (iv) No.
6.6(a) Moment of applied force about hinge axis is required. This is the product of force and perpendicular distance. So, if the distance is reduced then the force must be increased for the same
effect.
(b)Friction opposes motion and is proportional to the normal reaction between the object and the ground. This reaction is larger than pushing.
(c)Circular motion has force towards the centre which increases as the square of the angular velocity.
(d)As capsule moves in an approximately circular orbit, the astronaut experiences a balance between gravitation and centrifugal force.
The differential equation has the form
6.7 
where x is the vertical car body displacement, y is the ramp profile, λ is the spring constant, k is the damper constant, n is the speed of the car at time t after the ramp is hit and m is the mass of the car.
For case A, the motion can be described by the equations
and
6.8
where T is the force between the arm and the door, G is the spring force, r = PB, b = BS, l = PS, θ is the angle at which the door is open, π = BSP and I is the moment of inertia of the door. [case B – no solution supplied]
Variable mass problem. The equations of motion are
6.9 
and
where x is the length of vertical chain, l the total chain length, T the tension in the chain at the deck edge, ρ the mass per unit length of the chain, μ the coefficient of friction and v the velocity.