Physics about Newton's Law

Two trucks collide, P truck stopped (mass = 5500kg), Q truck (mass=3000kg) hit P. Then both trucks moved forward 30m after collision
Friction acting on trucks was 6000N
a) What was the speed of the trucks after collision
兩車一起滑行時的加速度
F = (m + M)a
6000 = (5500 + 3000) a
a = 0.706ms-2
利用等加速運動公式
v2 = u2 + 2as
0 = u2 + 2(-0.706)(30)
u = 6.51ms-1

b)If the time of collision was0.05S. what was the force acting on truck P
P車被撞時的平均加速度
v = u + at
6.51 = 0 + a(0.5)
a = 13.0ms-2
碰撞時平均受力
F = ma
= 5500x13
= 71500N

c) what was the force acting on truck Q?
依牛頓第三定律(作用力等於反作用力)所以Q
所受的力亦為
71500N (方向和P所受的相反)

d) find the speed of truck Q before collision
Q車的加速度。
F = ma
71500 = 3000a
a = 23.8ms-2
所以Q車的初速為u
v = u + at
6.51 = u + (-23.8)(0.5)
u = 18.4ms-1

F=ma , it is rule