Ng 發問於 科學及數學其他 - 科學 · 1 十年前

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

b)If the time of collision was0.05S. what was the force acting on truck P

c) what was the force acting on truck Q?

d) find the speed of truck Q before collision

Would you like also explain the friction as the F of the equation: F=ma ?

I understand the "m" but still can't get the point about F

### 2 個解答

• ?
Lv 7
1 十年前
最愛解答

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

• 1 十年前

F=ma , it is rule