Physics-Mechanics問題

1.a. By Newton’s 2nd law of motion
F = (mp +mq)a
-6000 = (5500 + 3000)a
Acceleration, a =-0.706 ms-2
By v2 = u2+ 2as
0 = u2 +2(-0.706)(30)
Speed of trucks aftercollision, u = 6.51 ms-1

b. By impulse =change of momentum
Ft = mp(v- u)
F(0.05) = (5500)(0 –6.51)
Force acting on truckP, F = -7.16 X 105 N
Takeing themagnitude, F = 7.16 X 105 N

c. By Newton’s 3rd law of motion, the force actingon truck Q and that of P are action and reaction pair. So, they are equal inmagnitude but opposite in direction.
So, force acting on Q =7.16 X 105 N

d. By the law ofconservation of momentum
mqu = (mp+ mq)v
(3000)u = (5500 +3000)(6.51)
Speed of truck Qbefore collision, u = 18.4 ms-1

2.i. By s = ut + 1/2at2
0.2 = 0 + 1/2 (10)t2
Reaction time, t =0.2 s

ii. The result willnot be affected. It is because the two rulers share the same acceleration dueto gravity. Therefore, assume the reaction times of John be the same in the twocases, the two rulers must have travelled the same distance in this timeinterval.

iii. The graduatedscale on the ruler is not correct.
Take 5 cm be anexample.
By s = ut + 1/2 at2
0.05 = 0 + 1/2 (10)t2
Reaction time, t = 0.1s, but not 0.05 s

(a)Find the speed of the trucks after the collision
conservation of energy,
all K.E. lost to workdone against friction
0.5mv^2 = Friction x s
0.5(5500+3000)v^2 = 6000x30
v = 6.5ms^-1

(b)If time of collision was 0.05 s, find the force acting on truck P.
F = m(v - u)/t
= 5500x6.5 / 0.05
= 715000N

(c)Hence, find the force acting on truck Q.
by Newton's III law, action and reaction pairs
it is also 715000N but in opposite direction.

(d)Find the speed of truck Q before collision.
again, Ft = m(v-u)
-715000x0.05 = 3000(6.5 -u)
u = 18.4 ms^-1

2.(a) u = 0 ms^-1 s = ut + 0.5at^2
s = 0.2 m 0.2 = 0 + 5t^2
a = 10 ms^-2 t = √0.04
= 0.2 s

(b)nothing change, because gravitational acceleration is independent of mass.