This question is not complete, the complete question is;
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 170 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 22.0 kg is thrown sideways out of the car with a speed of 2.50 m/s relative to the car's initial velocity.
Part B
An object with a mass of 22.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.
Answer:
Part A) the final velocity of the car is 4.6 m/s
Part B) the final velocity of the car is 5.28 m/s
Explanation:
Given the data in the question;
Total mass (m₁+m₂) = 170 kg
velocity of magnitude Vx = 4.60 m/s
PART A)
An object with a mass of 22.0 kg is thrown sideways out of the car with a speed of 2.50 m/s relative to the car's initial velocity,
i.e
m₂ = 22.0 kg
so m₁ = 170 - 22 = 148 kg
so, we apply conservation of momentum
since the object thrown out of the car, it has nothing to do with the car's velocity.
(m₁+m₂)Vx = m₁Vx₁ + m₂Vx₂
we substitute
(170)4.60 = 148Vx₁ + 22(4.60)
782 = 148Vx₁ + 101.2
148Vx₁ = 782 - 101.2
148Vx₁ = 680.8
Vx₁ = 680.8 / 148
Vx₁ = 4.6 m/s
Therefore, the final velocity of the car is 4.6 m/s
Part B)
An object with a mass of 22.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.
Vx = V(m₁+m₂) / ((m₁+m₂) - m₁)
we substitute
Vx = 4.60(170) / ((170) - 22)
Vx = 782 / 148
Vx = 5.28 m/s
Therefore, the final velocity of the car is 5.28 m/s
