4) a truck moving at 35.0 mph collides head on with a car traveling in the opposite direction. the mass of the car is 550 kg, the mass of the truck is 2200 kg. during the collision both vehicles come to a stop in 0.150 s as the front end of both vehicles’ crumples. what was the acceleration of the car? assume that during the collision the acceleration of each vehicle is constant. also, assume that frictional force are small compared to the forces compared to the contact forces between the vehicles during the colision.

The principle of conservation of linear momentum states that total momentum between colliding bodies before and after collision is constant. In other words, if two bodies A and B collide, then the sum of their initial momenta is equal to the sum of their final momenta.

From the question, the two bodies are the truck and the car.

Let the momentum of the truck before collision be [tex]m_{t}[/tex] x [tex]u_{t}[/tex]

Let the momentum of the car before collision be [tex]m_{c}[/tex] x [tex]u_{c}[/tex]

Let the momentum of the truck before collision be [tex]m_{t}[/tex] x [tex]v_{t}[/tex]

Let the momentum of the car before collision be [tex]m_{c}[/tex] x [tex]v_{c}[/tex]

From the principle of linear momentum, we have that;

[tex]m_{t}[/tex] x [tex]u_{t}[/tex] + [tex]m_{c}[/tex] x [tex]u_{c}[/tex] = [tex]m_{t}[/tex] x [tex]v_{t}[/tex] + [tex]m_{c}[/tex] x [tex]v_{c}[/tex] ---------------(i)

Where;

[tex]m_{t}[/tex] = mass of the truck = 2200kg

[tex]u_{t}[/tex] = initial velocity of the truck before collision = 35.0mph = 35.0 x 0.45m/s = 15.75m/s

[tex]m_{c}[/tex] = mass of the car = 550kg

[tex]u_{c}[/tex] = initial velocity of the car before collision = -u [since the car is travelling in the opposite direction to the truck]

[tex]v_{t}[/tex] = velocity of the truck after collision = 0 [since the truck comes to a stop]

[tex]v_{c}[/tex] = velocity of the car after collision = 0 [since the car comes to a stop]

Substitute these values into equation (i) as follows;

(2200 x 15.75) + (550 x (-u)) = (2200 x 0 + 550 x 0)

(2200 x 15.75) + (550 x (-u)) = 0

34650 - 550u = 0

550u = 34650

u = 63m/s

Therefore, the initial velocity of the car is 63m/s

Now to calculate the acceleration of the car, we use one of the equations of motion as follows;

v = u + at --------------(ii)

Where;

u = initial velocity of the car = 63m/s

v = final velocity of the car = 0 [since the car comes to a stop after collision]

a = acceleration of the car

t = time taken for the car to stop = 0.150s

Substitute these values into equation (ii) as follows;

0 = 63 + a(0.15)

0.15a = -63

a = - [tex]\frac{63}{0.15}[/tex]

a = - 420m/s²

Therefore, the acceleration of the car was -420m/s². The negative sign shows that the car was actually decelerating.