Session 2: Revision Practice Solutions

Session 2: Practice Questions Measurements

V  p r4  1) Given t 8l n

,

η = 1.137  10-3

 p r4 t = 8l V  p r4 1 = 8l  V     t   V   p r l t  = +4 + +   p r l  V    t  = 0.02 + 4  0.05  + 0.005 + 0.03 

= 0.255



 =  0.255  1.137  10 -3



= 0.3  10 -3 (1 s.f.)   =  1.1  0.3   10 -3 kgm -1 s -1

Kinematics

1(a) Velocity is the rate of change of displacement; Acceleration is the rate of change of velocity

a=

v-u t



from definition 

 v = u  at

------- (1)

Displacement s is the area under the v-t graph. t

 v dt =  (u + at) dt

Hence, s =

0

t

0

1 2 at ------- (2) 2 Another method is to use the area of a trapezium under the graph (in your lecture notes) = ut +

(b) From (1), t =

v-u a

Sub this into (2) 2

1  v- u  v -u   + a  2  a   a 

s=u 



 2as = 2uv - 2u 2  v 2 - 2uv + u 2  v 2  u 2 + 2as Condition: acceleration is constant

2)(i) Vv = 15 sin60˚ = 13 ms-1 VH = 15 cos 60˚ = 7.5 ms-1 (ii) 1. v2 = u2 + 2as 02 = 132 + 2(-9.81)s s = 8.61 m



2. 1 s  ut  gt 2 2 1 0  13t  (9.81)t 2 2  t  0 or t  2.65s 3. Horizontal distance = 7.5  2.65 = 19.9 m (iii)

8.61 m

N

A

t = 2..65 s

t = 0s 19.9 m For path A, the sketch must show - shorter range - lower maximum height - point of maximum height displaced to the left - Asymmetrical path Possible explanations:

Shorter range: In the horizontal direction, the ball experiences a net deceleration, therefore the horizontal speed decreases with time thus resulting in a shorter range for path A. Lower max height: As the ball rises, it has to do work against air resistance, hence mechanical energy is used to overcome this resistance and the maximum height reached will be lower Max height displaced to the left: As the ball rises, it experiences a net deceleration that is larger than 9.81 ms-2 because air resistance is acting in the same direction as weight. Hence its vertical speed reduces to zero in a shorter interval of time. As a result it reaches its maximum height sooner. This explains why the maximum height is displaced to the left. Asymmetric path: The path is asymmetric because horizontal speed is not uniform (horizontal distance covered becomes less and less with time) and in the vertical direction, the vertical acceleration experienced by the ball is not constant.