Guys, can you please answer these three questions. I'm having trouble doing them.

17) a) 30 kg·m/s

b) 30 kg·m/s

c) 0.3 m/s.

18) 2.4 m/s

19) 360,000 N

The different make of  car is safer because the force acting on it  is 28,800 N, resulting much less impact

Explanation:

17) By conservation of linear momentum principle, we have;

m₁·v₁ + m₂·v₂ = m₁·v₁' + m₂·v₂'

Where;

m₁ = The mass of the astronaut = 100 kg

v₁ = The initial velocity of the astronaut = 0 m/s

m₂ = The mass of the equipment box = 2.5 kg

v₂ = The initial velocity of the equipment box = 0 m/s

v₁' = The final velocity of the astronaut

v₂' = The final velocity of the equipment box = 12 m/s

The final momentum of the equipment box = Mass of the equipment box × Final velocity

a) The final momentum of the equipment box = 2.5 kg × 12 m/s = 30 kg·m/s

b) Given that the box and the astronaut will move in opposite directions, we have;

m₁·0 + m₂·0 = -m₁·v₁' + m₂·v₂'

m₁·v₁' = m₂·v₂'

Therefore, the momentum of the astronaut will be equal to the momentum of the equipment box = 30 kg·m/s

c)  The velocity of the astronaut is given by th following relation;

100×0 + 2.5×0 = -100×v₁' + 2.5×12

100×v₁' = 2.5×12

v₁' = 2.5×12/100 = 30/100 = 0.3 m/s

The final velocity of the astronaut = 0.3 m/s.

18) By conservation of linear momentum principle, we have;

m₁·v₁ + m₂·v₂ = m₁·v₁' + m₂·v₂'

Where;

m₁ = The mass of the railway truck= 4000 kg

v₁ = The initial velocity of the railway truck= 8 m/s

m₂ = The mass of the stationary truck = 6000 kg

v₂ = The initial velocity of the stationary truck = 0 m/s

v₁' = The velocity of the two trucks immediately after collision

Therefore, we have;

4000 kg × 6 m/s + 6000 × 0 = (6000 + 4000) × v₁' = 10,000·v₁'

24,000 kg·m/s = 10,000·v₁'

v₁' = 24,000/10,000 = 2.4 m/s

The velocity of the two trucks immediately after collision is = 2.4 m/s

19) The change in kinetic energy of the car due to the collision = 1/2·m·(v₂² - v₁²)

The initial velocity of the car, v₁ = 12 m/s

The final velocity, v₂ = 0 m/s

The mass of the car, m= 600 kg

The time in which the car is brought to rest = 2 seconds

Initial time at point of collision, t₁ = 0 s

Time after collision, t₂ = 0.02 s

By Newton's second law of motion, we have

The force F acting on the car = Rate of change in momentum produced by the force

F = m × dv/dt

Force  = m × (v₂ - v₁)/(t₂ - t₁) = 600×(0-12)/(0.02 - 0) = -360,000 kg·m/s²

The force acting on the car  is equal and opposite to the force of the car =  360,000 N

b) For the different make of the car, we have;

m × (v₂ - v₁)/(t₂ - t₁) = 600×(0-12)/(0.25 - 0) = -28,800 kg·m/s². = -28,800 N

Therefore, the different make of  car is safer because the force acting on it  is 28,800 N, resulting much less impact.