Question

Two train cars moving in the same direction are going to be coupled together. The mass of the first car is 5,000 kg and is moving at 5 m/s; the second car weighs the same, but is moving at 1 m/s. How fast will the two coupled cars move and how much kinetic energy does the system lose from coupling the cars together after they collide

Answer 1

The two coupled cars will move at a final velocity of 3 m/s. The system loses 20,000 Joules of kinetic energy from coupling the cars together.

Step-by-step explanation:

The final velocity of the two coupled cars can be found using the conservation of momentum principle. The total momentum before the collision is equal to the total momentum after the collision.

To find the final velocity, we can use the formula:

m1v1 + m2v2 = (m1 + m2)v

where m1 and m2 are the masses of the cars, v1 and v2 are their initial velocities, and v is the final velocity.

Using the given values, we have:

(5000 kg)(5 m/s) + (5000 kg)(1 m/s) = (10000 kg)v

Simplifying the equation gives:

25000 kg m/s + 5000 kg m/s = 10000 kg v

30000 kg m/s = 10000 kg v

Dividing both sides by 10000 kg gives:

v = 3 m/s

Therefore, the two coupled cars will move at a final velocity of 3 m/s.

The amount of kinetic energy lost from coupling the cars together after they collide can be calculated using the formula:

KE lost = 1/2 m1(v1^2) + 1/2 m2(v2^2) - 1/2 (m1 + m2)(v^2)

Substituting the given values:

KE lost = 1/2 (5000 kg)(5 m/s)^2 + 1/2 (5000 kg)(1 m/s)^2 - 1/2 (10000 kg)(3 m/s)^2

KE lost = 1/2 (5000 kg)(25 m^2/s^2) + 1/2 (5000 kg)(1 m^2/s^2) - 1/2 (10000 kg)(9 m^2/s^2)

Simplifying the equation gives:

KE lost = 62500 kg m^2/s^2 + 2500 kg m^2/s^2 - 45000 kg m^2/s^2

KE lost = 20000 kg m^2/s^2

Therefore, the system loses 20,000 Joules of kinetic energy from coupling the cars together.

Answer 2

Answer:
`3\ m/s,\ 20,00\ J`

Step-by-step explanation:

Given

Mass of the first car
`m_1=5000\ kg`

Mass of the second car
`m_2=5000\ kg`

The velocity of the first car is
`v_1=5\ m/s`

The velocity of the second car is
`v_2=1\ m/s`

Conserving momentum, take
`v_o` as the velocity after coupling

`\Rightarrow m_1v_1+m_2v_2=\left( m_1+m_2\right)v_o\\\Rightarrow 5000* 5+5000* 1=\left( 10,000\right)v_o\\\\\Rightarrow v_o=(25,000+5000)/(10,000)\\\\\Rightarrow v_o=(30,000)/(10,000)\\\Rightarrow v_o=3\ m/s`

`\text{Initial kinetic Energy }K_1=(1)/(2)m_1v_1^2+(1)/(2)m_2v_2^2\\\\\Rightarrow K_1=(1)/(2)* 5000\left( 5^2+1^2\right)\\\\\Rightarrow K_1=65,000\ J\\\\\\\text{Final Kinetic Energy}\ K_2=(1)/(2)\left(m_1+m_2\right)v_o^2\\\\\Rightarrow K_2=(1)/(2)* 10,000* 3^2\\\\\Rightarrow K_2=45,000\ J\\\\\text{Kinetic energy lost is equivalent to change in Initial and final energy i.e.}\\\\\Rightarrow K_1-K_2=65,000-45,000\\\\\Rightarrow K_1-K_2=20,000\ J`