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) To determine if 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 energy is lost to other forms, such as heat or deformation.
In this scenario, the cars remain locked together after the collision, indicating that some energy has been lost. Therefore, this is an inelastic collision.
b) To calculate the velocity of the two cars immediately after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Before the collision:
Car 1 (stationary car) mass (m1) = 1500 kg, velocity (v1) = 0 m/s
Car 2 (moving car) mass (m2) = 1300 kg, velocity (v2) = 40 km/h = 11.11 m/s
After the collision:
Let the final velocity of both cars be V.
Total momentum before the collision = Total momentum after the collision
(m1 * v1) + (m2 * v2) = (m1 + m2) * V
(1500 kg * 0 m/s) + (1300 kg * 11.11 m/s) = (1500 kg + 1300 kg) * V
0 + 14443 kg*m/s = 2800 kg * V
V = 14443 kg*m/s / 2800 kg
V ≈ 5.16 m/s
Therefore, the velocity of the two cars immediately after the collision is approximately 5.16 m/s.
c) To calculate the deceleration, time taken to come to rest, and the distance traveled by the cars, we can use the equations of motion.
Deceleration (a) is given by:
a = (v - u) / t
Where:
v = final velocity = 0 m/s (since the cars come to rest)
u = initial velocity = 5.16 m/s (velocity immediately after the collision)
t = time taken to come to rest
Assuming the deceleration is constant and using the equation of motion:
v^2 = u^2 + 2 * a * s
Where:
v = final velocity = 0 m/s
u = initial velocity = 5.16 m/s
a = deceleration
s = distance traveled
Plugging in the known values, we can solve for the deceleration (a):
0^2 = (5.16 m/s)^2 + 2 * a * s
0 = 26.58 m^2/s^2 + 2 * a * s
Rearranging the equation:
2 * a * s = -26.58 m^2/s^2
Now, we can use the equation a = (v - u) / t to find the time taken to come to rest:
0 = (5.16 m/s - 0 m/s) / t
0 = 5.16 m/s / t
Rewriting the equation:
t = 5.16 m/s / 0
t = undefined
Since the denominator is zero, we cannot determine the exact time taken to come to rest.
Lastly, to find the distance traveled (s), we can use the equation of motion:
v^2 = u^2 + 2 * a * s
Plugging in the known values:
0^2 = (5.16 m/s)^2 + 2 * a * s
0 = 26.58 m^2/s^2 + 2 * a * s
Rearranging the equation:
2 * a * s = -26.58 m^2/s^2
Substituting the value of the deceleration (a) derived earlier:
-26.58 m^2/s^2 = 2 * (-26.58 m^2/s^2) * s
Simplifying:
-26.58 m^2/s^2 = -53.16 m^2/s^2 * s
Dividing both sides by -53.16 m^2/s^2:
s = -26.58 m^2/s^2 / -53.16 m^2/s^2
s ≈ 0.5 m
Therefore, the deceleration of the cars is approximately -26.58 m^2/s^2 (negative sign indicates deceleration), the time taken to come to rest is undefined, and the distance traveled by the cars is approximately 0.5 m.