An airplane flies east at a speed of 120 m/s the airplane has a mass of 3200
1 answer:
1) The momentum of the airplane is east
2) The impulse applied to the airplane is 32,000 kg m/s west
Explanation:
1)
The momentum of an object is defined as
where
m is the mass of the object
v is its velocity
The airplane in this problem has
m = 3200 kg is its mass
v = 120 m/s east is its velocity
So, its momentum is
And since momentum is a vector, it also has a direction, which is the same as the velocity (east).
2)
The impulse applied to an object is equal to the change in momentum of the object:
where
m is the mass
v is the final velocity
u is the initial velocity
For the airplane in this problem,
m = 3200 kg
u = 120 m/s
v = 110 m/s
So, the impulse is
And the negative sign means the direction of the impulse is opposite to the airplane motion, so west.
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The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.
<h3>What is the law of conservation of momentum?</h3>According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
(m₁) is the mass of hockey player 1= 91.0-kg
(m₂) is the mass of hockey player 2= 91.0-kg
(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.
(u₂) is the velocity before the collision of hockey player 2=?
a)
Momentum before the collision;
Momentum before the collision = 1237.6 kg m/s².
b)
The velocity of the two hockey players after the collision from the law of conservation of the momentum as:
Momentum before collision = Momentum after the collision
1237.6 kg m/s² = (m₁+m₂)V
1237.6 kg m/s² =(2 ×91.0-kg )V
V=6.8 m/sec.
Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.
To learn more about the law of conservation of momentum refer;
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