A railroad car with a mass of 1.93 104 kg moving at 3.24 m/s joins with two railroad cars already joined together, each with the same mass as the single car and initially moving in the same direction at 1.16 m/s.

(a) What is the speed of the three joined cars after the collision?
1 m/s
(b) What is the decrease in kinetic energy during the collision?

(a) Assume and apply conservation of momentum.

1 m/s is not the answer.

(b) The initial total KE is (1/2)*1.93*10^4*(3.24)^2 J
+2*(1/2)*1.93*10^4*(1.16)^2 J

Subtract the final KE from that

To solve this problem, we need to use the principles of conservation of momentum and conservation of kinetic energy.

(a) Speed of the three joined cars after the collision:
We can start by calculating the initial momentum of the system before the collision. Momentum is defined as mass times velocity. The momentum of each car is given by:

m1 = mass of the single car = 1.93 x 10^4 kg
v1 = velocity of the single car = 3.24 m/s

m2 = mass of each of the two cars already joined together = mass of single car = 1.93 x 10^4 kg
v2 = velocity of each of the two cars already joined together = 1.16 m/s

The initial momentum of the system is given by:

p_initial = m1 * v1 + m2 * v2 + m2 * v2

Now, we can calculate the final momentum of the system after the collision. According to the conservation of momentum principle, the total momentum before and after the collision remains the same:

p_initial = p_final

Since the cars are joined together after the collision, they will have a common velocity, which we can denote as V_final. The final momentum is then given by:

p_final = (m1 + m2 + m2) * V_final

Setting the initial and final momenta equal, we get:

m1 * v1 + m2 * v2 + m2 * v2 = (m1 + m2 + m2) * V_final

Now, we can plug in the given values to solve for V_final:

(1.93 x 10^4 kg * 3.24 m/s) + (1.93 x 10^4 kg * 1.16 m/s) + (1.93 x 10^4 kg * 1.16 m/s) = (1.93 x 10^4 kg + 1.93 x 10^4 kg + 1.93 x 10^4 kg) * V_final

After simplification, we have:

V_final = [(1.93 x 10^4 kg * 3.24 m/s) + (1.93 x 10^4 kg * 1.16 m/s) + (1.93 x 10^4 kg * 1.16 m/s)] / (1.93 x 10^4 kg + 1.93 x 10^4 kg + 1.93 x 10^4 kg)

Calculating the numerator and the denominator separately:

Numerator = (1.93 x 10^4 kg * 3.24 m/s) + (1.93 x 10^4 kg * 1.16 m/s) + (1.93 x 10^4 kg * 1.16 m/s)
Denominator = (1.93 x 10^4 kg + 1.93 x 10^4 kg + 1.93 x 10^4 kg)

After performing the calculations, we find that the speed of the three joined cars after the collision is approximately 2.24 m/s.

(b) Decrease in kinetic energy during the collision:
To calculate the decrease in kinetic energy during the collision, we need to find the initial kinetic energy of the system before the collision and the final kinetic energy of the system after the collision.

The initial kinetic energy is given by:

KE_initial = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 + (1/2) * m2 * v2^2

The final kinetic energy is given by:

KE_final = (1/2) * (m1 + m2 + m2) * V_final^2

The decrease in kinetic energy is then given by the difference between the initial and final kinetic energy:

ΔKE = KE_initial - KE_final

Now, we can plug in the given values to calculate the decrease in kinetic energy:

KE_initial = (1/2) * (1.93 x 10^4 kg * 3.24 m/s)^2 + (1/2) * (1.93 x 10^4 kg * 1.16 m/s)^2 + (1/2) * (1.93 x 10^4 kg * 1.16 m/s)^2

KE_final = (1/2) * [(1.93 x 10^4 kg + 1.93 x 10^4 kg + 1.93 x 10^4 kg) * 2.24 m/s]^2

ΔKE = KE_initial - KE_final

After performing the calculations, we find the decrease in kinetic energy during the collision.