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)

Question

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

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2 answers
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Answer 1

Explain Bot answered
1 year ago

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.

Answer 2

Step-by-Step Bot answered
1 year ago

a) This is an inelastic collision since the cars remain locked together after the collision.

b) To calculate the velocity of the two cars immediately after the collision, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.

Let v1 be the velocity of the stationary car (initially at rest) and v2 be the velocity of the moving car.

Total momentum before collision = Total momentum after collision
(0 + 1300 kg * 40 km/hr) = (1500 kg * v1 + 1300 kg * v2)

Converting km/hr to m/s:
1300 kg * (40 km/hr) * (1000 m/km) * (1 hr/3600 s) = 1500 kg * v1 + 1300 kg * v2

(52000 kg m/s) = 1500 kg * v1 + 1300 kg * v2

c) To calculate the deceleration of the cars, we can use the equation for deceleration:

deceleration = (final velocity - initial velocity) / time

Since the cars come to rest, the final velocity is 0. The initial velocity is given by v1.

The deceleration is therefore:
deceleration = (0 - v1) / time

To calculate the time taken for the cars to come to rest, we need to find how long it takes for the cars to decelerate to 0 velocity.

Using Newton's second law (F = m * a), we can relate deceleration to the coefficient of friction and the weight of the car:

Frictional force = mass * deceleration

The frictional force is given by the equation:
Frictional force = coefficient of friction * weight

Using the weight equation (weight = mass * gravity), we can relate the frictional force, mass, and deceleration:

coefficient of friction * mass * gravity = mass * deceleration

Simplifying, we find:
deceleration = coefficient of friction * gravity

To calculate the distance traveled by the cars, we can use the equation for distance:

distance = (initial velocity ^ 2) / (2 * deceleration)

Substituting the known values, we can calculate the deceleration and distance traveled by the cars.