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A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car initially has a total mass (car and contents) of 150 kg and is traveling east with a velocity of magnitude 4.60 m/s. Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks.

Part A : An object with a mass of 21.0 kg is thrown sideways out of the car with a speed of 2.40 m/s relative to the car's initial velocity.
Part B : An object with a mass of 21.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.
Part C : An object with a mass of 21.0 kg is thrown into the car with a velocity of 6.00 m/s relative to the ground and opposite in direction to the initial velocity of the car.

Respuesta :

Answer:

Part a)

[tex]v = 4.60 m/s[/tex]

Part b)

[tex]v_{1f} = 5.24 m/s[/tex]

Part c)

[tex]v = 3.3 m/s[/tex]

Explanation:

Part a)

Object is thrown sideways

since the railroad car is moving on the track so there is no change in the momentum in the direction of motion of the car

so final speed of the car will not change

[tex]v = 4.60 m/s[/tex]

Part b)

By momentum conservation we can write

[tex]m_1v_{1i} = m_1v_{1f} + m_2v_{2f}[/tex]

[tex]150(4.60) = 150 v_{1f} - 21(4.60)[/tex]

[tex]v_{1f} = 5.24 m/s[/tex]

Part c)

Again by momentum conservation we have

[tex]m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f[/tex]

[tex]150(4.60) - 21(6.00) = (150 + 21) v[/tex]

[tex]v = 3.3 m/s[/tex]

a. The final velocity of the car is 4.60 m/s

b. The final velocity of the car is 5.24 m/s

c. The final velocity of the car is 3.3 m/s

Calculation of the final velocity of each case should be:

a.

Since the object is thrown sideways.

Also, the railroad car is moving on the track due to this there is no change in the momentum in the direction.

Therefore, the final speed of the car will not change so it should be 4.60 m/s

b. Here we applied conservation of momentum

[tex]m_1v_1i = m_1v_1f + m_2v_2f\\\\150(4.60) = 150v_1f+ m_2v_2f\\\\v_1f= 5.24 m/s[/tex]

c. Here we applied conservation of momentum

[tex]m_1v_1i + m_2v_2i = (m_1 + m_2)v_f\\\\150(4.60) - 21(6.00) = (150 + 21)v\\\\v = 3.3 m/s[/tex]

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