A1 MS.docx MARKING SCHEME FOEA1 QUESTIONS

KINEMATICS A1

1. (0) The extra time the time latest blue bal to reach ts maximum height an return tte
lunch point. takes 030-108 >t 3.05 to gett the topo T= 6s

o

(} Let rbe the time at which decaeration tarts and Tthe required time for the entire motion. The

deceleration LA 80, hence 7=£=30. The 108

Area oTer=90. Hence,

(i straight Ine untl? followed by concave down parabola wth zero slope at + 64

y=0+30:8-x10:80' =-80m. Teele 20m

2. m 20-10:802-50ms

(U) The height above the te 0-30 O 45m. The distance traveled

15 45145180 170m so the average sped ls
(e) Lette na veocky be u. Then at ¢=2 the velocity =u~10%20 andat¢= 4

ve =U=10%40. Me have that y, =, £0 u-10x20--W-10x40) or 20-60 £0 430m

aivelociy component is y, 20-10x20=-20m5", The angle with the horizontals

y=012+254n19 204654 1108204654" -285m. Bottom of bulls at het of

285-0.12- 273 ma0 the ball goes over the crossbar. Something ike Roberto Bags

3. (At the lowest pont the resultant forces T=mg and so T-mg = e Ris the length of

the string By energy conservation: maß = Em => mv! =2mgR, Henc

ofresoutionsis 398240
Dan

= srods

207 LE 9373.032m
707 750

16) From M atx yla la . SHE rhen the required ratios

(62 Atthe top point onthe rack the resultant forces Ns mg and N

20H =4GR eV >! =2gH SR. Hence ph = 29-49 =H,

4. Tsind amg and Teo Le the reutant fo

tensions the hypotenuse of a igh onged tran

ET

(the angles found from: 0-15)

6. rc proportonalto speed Stokes wil ae on the body epposng the motion
force wi neeace scent otra the ne fore on te body

v 205010), 5.9, 2300-1100
6416

1 There when the largest and amas possible mare a

7. Ke energy before the o

Lao,

8. “The acelration increases because the force the same but the mass decreaces
‘The sceieration becomes zero when al the felis used up.
(One panic welcty makes an ange with the orgnalincident partie direction. The othe

Apalyng conserition af mementum song the erection ofthe incident pace and
angerto we fe:

ne
ova sone
Omen mn Arte
Bernd weed

We have two equations fortwo unknowns:

u=ves6+winé
O-vaind-wens®

From the second: wavtand. Subaitingin the fe

Uaveossevtandilnd end) Hence, v=ucosó.

Then, wavtand =ucorStand using

Tenia Kinetic energy Lut. The Sialic

Much +

Se, the clon east

9. (Forthoze we know vectors, there ira much simpler olution: et be the inital

momentum, and À, À, the moments ofthe two particles after the coliion. Then
‘Thus, aking dot product:

PPB +B BAIA AA

Pp + pl Since BO because the vectors are perpendicular

Tu 2 and kinetic energy is conserve)

10. Practice exam-style questions
A.1 Kinematics.
Papert

Paper 2
1=022: (or show calculation i fl)
Straight line up rom 2er to r= 0.
with gradient of
Time to reach net
Distance fallen = 0.064 m
0064 < (024-015) m

Usev=u rat

Truck has constant sped (horizontal line) of 15 a +

Car stare at origin and ives a a tsiht line to 24m 0 at 10
Froa 10 à t 15 scars speed is shown by a horizontal Kine
Afr 625 +

‘After 10 the caris 30 behind the tl

‘Another 3.3 «before they have travelled the came distance,
Tota time