a stationary car is hit from behind by another car travelling at 40km per hr. After collision both cars remain locked together. The masses of the stationary car and the moving car are 1500kg and 1300kg respectively (use g=9.8ms-2)
a) is this an elastic or inelastic collision
b) calculate the velocity of the two cars immediatly after the collision
c) If the brakes of the stationary cars are applied before impact and the coefficient of friction between the wheels and the road surface is 0.4 calculate the decelaration of the cars.the time taken for the cars to come to rest and the distance travelled by the cars
a) This is the inelastic collision.
b)
v =40 km/h=11.11 m/s
M(st)•0+M(mov) •v=(M(st)+M(mov)) •V
V= M(mov) •v/(M(st)+M(mov))=1300•11.11/(1300+1500) =5.158 m/s.
c) m•a =F(fr) = k•m•g
a=k•g =0.4•9.8 =3.92 m/s.
s=v^2/2a = (11.11)^2/2•3.92 = 15.72 m
0 = v- at
t = v/a – 11.11/3.92 = 2.83 s.
a) To determine whether the collision is elastic or inelastic, we need to consider whether kinetic energy is conserved during the collision. In an elastic collision, kinetic energy is conserved, meaning the total kinetic energy before the collision is equal to the total kinetic energy after the collision. In an inelastic collision, kinetic energy is not conserved, and some of the kinetic energy is transferred into other forms, such as heat or deformation.
b) To calculate the velocity of the two cars after the collision, we need to apply the principles of conservation of momentum. The momentum before the collision is equal to the momentum after the collision. The momentum of an object is given by the product of its mass and velocity.
Let's denote the velocity of the stationary car after the collision as V1', and the velocity of the moving car after the collision as V2'.
The momentum before the collision is:
Momentum1 = mass1 * 0 (since the stationary car has zero velocity initially) = 0
The momentum after the collision is:
Momentum1' = mass1 * V1'
Momentum2' = mass2 * V2'
Since the two cars remain locked together after the collision, their final velocity is the same:
V1' = V2'
We can set up an equation using the principle of conservation of momentum:
Momentum1 + Momentum2 = Momentum1' + Momentum2'
0 + (mass2 * 40) = (mass1 * V1') + (mass2 * V2')
Now we can substitute the given values:
0 + (1300 * 40) = (1500 * V1') + (1300 * V1')
52000 = 1500V1' + 1300V1'
c) To calculate the deceleration of the cars when the brakes are applied, we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s since the cars come to rest)
u = initial velocity (V1')
a = acceleration (deceleration)
s = distance travelled
Since we know the initial velocity (V1') and the final velocity (0), we can rearrange the equation to solve for the acceleration (a):
a = (v^2 - u^2) / (2s)
Given that the coefficient of friction (μ) is 0.4, we can use the equation:
F(friction) = μ * mass * g
where:
F(friction) = force of friction
mass = mass of the stationary car
g = acceleration due to gravity (9.8 m/s^2)
Since the force of friction is the same as the force causing deceleration, we can equate the two equations:
F(friction) = mass * a
We can now substitute the values:
(0.4 * 1500 * 9.8) = 1500 * a
Finally, we can rearrange the equation to solve for acceleration (a):
a = (0.4 * 1500 * 9.8) / 1500
Once we have the acceleration, we can use the equation of motion again to calculate the time taken for the cars to come to rest and the distance travelled:
v = u + at
Given that the final velocity (v) is 0 and the initial velocity (u) is V1', we can solve for time (t):
0 = V1' + a * t
Rearranging the equation:
t = -V1' / a
Using the negative value because the initial velocity is in the opposite direction of the deceleration.