a) To determine whether the collision is elastic or inelastic, we need to consider whether kinetic energy is conserved. In an elastic collision, kinetic energy is conserved, while in an inelastic collision, kinetic energy is not conserved.
In this case, the cars remain locked together after the collision, indicating that the collision is inelastic. This is because some energy is lost in deformations or other processes during the collision.
b) To calculate the velocity of the two cars immediately after the collision, we can use the principle of conservation of momentum. In an inelastic collision, the total momentum before the collision is equal to the total momentum after the collision.
Let's assume that the mass of the stationary car is m1 (1500 kg) and the mass of the moving car is m2 (1300 kg). The initial velocity of the moving car is 40 km/hr, which we need to convert to m/s for consistent units.
Converting 40 km/hr to m/s:
(40 km/hr) * (1000 m/km) * (1 hr/3600 s) = 11.11 m/s
Using the conservation of momentum:
(m1 * 0) + (m2 * 11.11 m/s) = (m1 + m2) * vf
Where vf is the final velocity of the combined cars. Since the stationary car is at rest initially (velocity = 0), we can simplify the equation:
(1300 kg * 11.11 m/s) = (1500 kg + 1300 kg) * vf
vf = (1300 kg * 11.11 m/s) / (2800 kg)
vf ≈ 5.15 m/s
Therefore, the velocity of the two cars immediately after the collision is approximately 5.15 m/s in the direction of the moving car before the collision.
c) To calculate the deceleration of the cars, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s as the cars come to rest)
u = initial velocity (5.15 m/s as calculated above)
a = deceleration (to be computed)
s = distance traveled
Rearranging the equation to solve for a:
a = (v^2 - u^2) / (2s)
Plugging in the values:
a = (0^2 - 5.15^2) / (2 * s)
a = -26.52 / (2 * s)
a = -13.26 / s
Given the coefficient of friction between the wheels and the road surface (μ = 0.4), we can also use the equation of motion for constant deceleration:
v^2 = u^2 + 2as
Since the final velocity is 0, we can rearrange the equation to solve for s:
s = (v^2 - u^2) / (2a)
Using the same values for v and u, and the coefficient of friction for a:
s = (0^2 - 5.15^2) / (2 * -0.4 * 9.8)
s ≈ 6.54 m
Therefore, the deceleration of the cars is approximately -13.26 m/s^2 (negative because it is in the opposite direction of motion), the time taken for the cars to come to rest is not calculated but can be determined using the final velocity and deceleration, and the distance traveled by the cars is approximately 6.54 meters.