JC2Raffles Institution (Junior College)2009

Physics Revision Tutorial1

Kinematics

Solutions

Rectilinear Motion:

1 / A





Equal areas under the graph above and below axes since distance travelled upward equals distance travelled downwards.
2 / A
Since y t 2 , distance fallen after 2 s will be (42) = 8 storeys. Ball will be 8th storey below the 10th storey. Hence ball will be at 2nd storey.

Projectile Motion:

7 / Let t be time to fall through vertical distance of 180 m.

Horizontal distance travelled = vxt = 500  6.058 = 3000 m (to 2 s.f.)

DISCUSSION QUESTIONS

Rectilinear Motion

9 / (a) / Thinking distance st = u tr since the car will be travelling at a constant speed u during the reaction time tr. Taking reaction time to be constant, thinking distance is therefore proportional to speed.
When the brakes are applied, the car experiences a negative acceleration a. Using
Braking distance
Braking distance is proportional to the square of the speed.
(b) /

(b) (c)

Speed / km h1 / Speed / m s1 / Thinking distance* / m / Braking distance** / m / a / m s2 / tr / s
40 / 11.1 / 6.7 / 9.4 / 6.57 / 0.603
50 / 13.9 / 8.3 / 14.6 / 6.61 / 0.597
60 / 16.7 / 10.0 / 21.0 / 6.61 / 0.600
70 / 19.4 / 11.7 / 28.6 / 6.61 / 0.603
80 / 22.2 / 13.3 / 37.4 / 6.59 / 0.600
Average = / 6.6 (2 s.f.) / 0.60 (2 s.f.)
(c) / To find reaction time, use for the data given and find average (see table above).
For u = ,
Overall distance
(d) / (i) / When the road is wet, adecreases in magnitude and hence sbwill be larger. Thinking distance remains unchanged.
(ii) / When the driver is not fully alert, reaction time is longer. Thinking distance is longer whilst the braking distance is unchanged.
(e) / Assuming that during the reaction time, the car moves with constant velocity down the slope,
During braking,
acceleration down the slope

Overall stopping distance = 10.0 + 28.36 = 38.4 m
10 / TYS Pg 43 Q12 (N97/II/1)

TYS Pg 45 Q17 (N03/III/1)

11 / (c) / (i) / Intercept on the t - axis = 1.76 s
(ii) / Acceleration can be found from the gradient of the tangent to the curve.
(iii) / The ball was projected with an upward velocity. Hence, just after the ball leaves the thrower’s hand, it experiences a downward drag force. This drag force is large when the velocity is high, which is the case at the instant the ball was projected. The total downward force is now the sum of the weight of the ball and the drag force on the ball. Hence the ball experiences a downward acceleration larger than the acceleration due to gravity.
(iv) / The acceleration is equal to g only when the ball is momentarily at rest at its maximum height. At that instant when t = 1.76 s, the drag force is zero and the only force acting on the ball is the gravitational force of the Earth on the ball
(v) /
(d) / On its way up, the drag force and the weight of the ball both act to slow down the ball. Hence the ball is brought to rest in a shorter time than if there is no drag force. On the way down, the drag force acts opposite in direction to the weight. The net accelerating force is smaller than if no drag force is acting. The average velocity of the ball on the way down is smaller than that on its way up. Hence the ball will take a longer time to cover the same distance downward.
(e) / (i) /
(ii) / On its way up, kinetic energy of the ball is converted to gravitational P.E. with some kinetic energy being used to overcome drag force (dissipated as heat). On the way down, the gravitational P.E. is now converted to kinetic energy of the ball with some of the gravitational P.E. also being used to overcome the drag force. The ball will return to its starting point with less kinetic energy than when it leaves the thrower’s hand.

Projectile Motion

12 /

Let t be time to fall through vertical distance of 500 m.

Horizontal distance travelled = vxt = 50  4.58 = 229 m (3 s.f.)
13 /

Horizontal speed

14 TYS Pg 46 Q19 (N2000/II/2)

(i) / 1. /
2. / Time taken to reach max height is ½ t. At max height, vy= 0.
Using vy = uy + ay t
0 = v sin  g (½ t)

(ii) /
Note to Teachers: TYS gives answer as 32o but Cambridge Examiners accept both.
(iii) / In practice, due to air resistance, the horizontal range would become smaller.
Based on , the horizontal range increases as increases up to a maximum when  = 45o, i.e. sin 2 = 1. Beyond  = 45o, the horizontal range decreases as  increases.
If  is originally 32o, the angle of projection must be increased.
If is originally 58o, the angle of projection must be decreased.
Note to Teachers: TYS gives the wrong answer. Cambridge examiners only mentioned greater than 32o.

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