Phy CE 2006 paper 2 #4

Ok. I know how to do it.

the car stopped because of the friction between the tyre and the road. This force provided deceleration and according to Newton's 2nd law, F=ma

Assume:

Mass of the car = m
Acceleration due to gravity = g = 10m/s^2
deceleration = a

1)
friction = 0.65 x weight = 0.65mg (N)
But by Newton's 2nd law,
friction = -ma (the negative sign means deceleration)

Hence, -ma = 0.65mg

a = -0.65g = - 6.5m/s^2 (negative sign means deceleration)

2)
Assuming the deceleration is uniform,

s = 22.3m
v = 0m/s (because finally the car stopped)
a = - 6.5m/s^2

By (v^2) - (u^2) = 2as,

we have:
- (u^2) = 2 (-6.5)(22.3)
u^2 = 289.9

u = 17.02m/s, this is the initial speed of the car when the car started to skid

Hence, answer is D

**Note that the units in the question are wrong. The correct one should be m/s

THanks for reading

比多個alternate method 你知道吓啦, 就係用conservation of energy.

因為架車本來係用緊speed v 行得好好, 但係當apply 左brake 之後就停左.

咁即係 work done against friction = kinetic energy in car.
Fxs = 0.5xmxv^2
0.65xmxg=0.5xmxv^2 (個m 唔使知, 因為可以約走左; take g=10)
0.65x10/0.5 = v^2
v= 17ms-1

正所謂條條大路通羅馬, 諗多d, 聽多d, 咁就可以學多d 啦.

let the mass be m,f be the frictional force, take g be 10m/s^2

weight=mg

friction=0.65[weight} (given)
　　　=0.65mg
Net force =ma (newton's second law of motion)
　 　　 f =ma
　0.65mg=ma
　　0.65g=a
　　　　a=-6.5m/s^2 (it is deceleration)
in this step,we've got s=22.3m a=-6.5m/s^2 and it is stoped, that v=0m/s
by using　　　v^2-u^2=2as
　　　　　　　0-u^2=2(-6.5)(22.3)
　　　　　　　　 u^2=289.9
　　　　　　　　　　u=17.0m/s
the anwer is D

個重點在於f=weight,唔係=mass
weight=mg

希望幫到你啦^^