a railroad handcar is moving along striaght, fricionless tracks. In each of the following cases the car initially has a total mass (car and contents of magniude 4.00 m/s^1. Find the final velocity of the car in each of the three cases.
A) a 40.0 kg mass is thown sideways out of the car with a velocity of magnitude 2.00 m/s^1 relative to the car.
B) a 40.0 kg mass is thrown backward (to the west)out of the car with a velocity of magnitude 4.00 m/s^1 relative to the initial motion of the car.
C) a 40.0 kg mass is thrown into the car with a velocity of the magnitude 6.00 m/s^1 relative to the ground and opposite in direction to the velocity of the car.
a) 4.00m/s
b)velocity of object is 0, so your answers should be (Mass of car * 4.00 )/(Mass of car - 40.0 kg)
c) 2 initial velocities this time, and the object is going the opposite direction. You should have [(Mass of Car * 4.00) - (40kg *6) ] / (Mass of car + 40kg)
a car of mass 500 is travelling at 24.a lion of 100kg drop onto the roof the car from the over handing branch show that the car will slow down to speed lot 20m/s.
help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee"""""""
To find the final velocity of the car in each case, we need to apply the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In this case, the railroad handcar is a closed system with no external forces acting on it, so the total momentum before and after each event should be the same.
Let's find the initial momentum, final momentum, and then use the principle of conservation of momentum to determine the final velocity of the car in each case.
A) A 40.0 kg mass is thrown sideways out of the car with a velocity of magnitude 2.00 m/s relative to the car.
Initial momentum (car + contents) = (mass of the car + mass of the contents) × initial velocity of the car = 4.00 kg × 0 m/s = 0 kg⋅m/s (because the car is at rest initially)
Final momentum (car + contents) = (mass of the car + mass of the contents) × final velocity of the car
According to the conservation of momentum, the initial momentum and final momentum must be equal. So we can set up an equation:
0 kg⋅m/s = (4.00 kg + 40.0 kg) × final velocity of the car
Simplifying the equation, we get:
0 = 44.0 kg × final velocity of the car
Since any number multiplied by zero is zero, it means that the final velocity of the car is zero (0 m/s).
B) A 40.0 kg mass is thrown backward (to the west) out of the car with a velocity of magnitude 4.00 m/s relative to the initial motion of the car.
In this case, we need to consider the direction of the mass being thrown. Since it is thrown backward (opposite to the initial motion of the car), the relative velocity of the mass with respect to the car would be the sum of the individual velocities (relative to the ground) in opposite directions.
Relative velocity of the mass with respect to the car = Velocity of the mass relative to the ground - Velocity of the car relative to the ground
Relative velocity of the mass with respect to the car = 4.00 m/s - 0 m/s = 4.00 m/s
Initial momentum (car + contents) = (mass of the car + mass of the contents) × initial velocity of the car = 4.00 kg × 0 m/s = 0 kg⋅m/s
Final momentum (car + contents) = (mass of the car + mass of the contents) × final velocity of the car
According to the conservation of momentum, the initial momentum and final momentum must be equal. So we can set up an equation:
0 kg⋅m/s = (4.00 kg + 40.0 kg) × final velocity of the car
Simplifying the equation, we get:
0 = 44.0 kg × final velocity of the car
Again, any number multiplied by zero is zero, so the final velocity of the car is zero (0 m/s).
C) A 40.0 kg mass is thrown into the car with a velocity of magnitude 6.00 m/s relative to the ground and opposite in direction to the velocity of the car.
Similar to case B, we need to consider the direction of the mass being thrown. Since it is thrown opposite to the velocity of the car, the relative velocity of the mass with respect to the car would be the difference of the individual velocities (relative to the ground) in opposite directions.
Relative velocity of the mass with respect to the car = Velocity of the mass relative to the ground - Velocity of the car relative to the ground
Relative velocity of the mass with respect to the car = 6.00 m/s - 0 m/s = 6.00 m/s
Initial momentum (car + contents) = (mass of the car + mass of the contents) × initial velocity of the car = 4.00 kg × 0 m/s = 0 kg⋅m/s
Final momentum (car + contents) = (mass of the car + mass of the contents) × final velocity of the car
According to the conservation of momentum, the initial momentum and final momentum must be equal. So we can set up an equation:
0 kg⋅m/s = (4.00 kg + 40.0 kg) × final velocity of the car
Simplifying the equation, we get:
0 = 44.0 kg × final velocity of the car
Again, any number multiplied by zero is zero, so the final velocity of the car is zero (0 m/s).
In all three cases, the final velocity of the car is zero.