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3 years ago

Why does the momentum before a collision equal the momentum after a collision?

Physics

2 answers:

Art [367]3 years ago

8 0

Answer: Correct option is C) because momentum is conserved

Explanation:

According to law of conservation of linear momentum if net external force on a system is zero then the total momentum of the system will be conserved .

For a collision occurring between object 1 and object 2 in an isolated system

If we consider the two colliding object as a single system then while collision the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

Thus momentum before collision equal the momentum after a collision because momentum is conserved .

kotykmax [81]3 years ago

3 0

The answer is C. Due to conservation of momentum.

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A bumblebee is flying to the right when a breeze causes the bee to slow down with a constant leftward acceleration of magnitude

romanna [79]

Answer:

3.75 m/s

Explanation:

Given that a bumblebee is flying to the right when a breeze causes the bee to slow down with a constant leftward acceleration of magnitude 0.50 m/s^2. After 2.0 s the bee is moving to the right with a speed of 2.75 m/ s

What was the velocity of the bumblebee right before the breeze?

Since the breeze blows with the acceleration of 0.5 m/s^2 for 2 seconds, we can calculate the magnitude of the velocity at which it moves.

Acceleration = velocity/ time

Substitute the acceleration and time into the formula

0.5 = V/2

Cross multiply

V = 2 × 0.5

V = 1 m/s

The bumblebee is travelling right ward while the wind travels leftward.

The relative velocity = 2.75 m/s

Let the bumblebee speed = Vb

While the wind = VW

Vb - Vw = 2.75

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3 years ago

In 1976, a 53 kg helicopter was built in Denmark. Suppose this helicopter flew east with a speed of 60.0 m/s and the total momen

Bumek [7]

Answer:

67 kg.

Explanation:

From the question,

Total momentum = (mass of the pilot+ mass of the helicopter)×velocity.

I = (m+M)v............................... Equation 1

Where I = Total momentum of the pilot and the helicopter, m = mass of the pilot, M = mass of the helicopter, v = velocity.

make m the subject of the equation

m = (I/v)-M.............................. Equation 2

Given: I = 7200 kg.m/s, v = 60 m/s, M = 53 kg.

Substitute these values into equation 2

m = (7200/60)-53

m = 120-53

m = 67 kg.

Hence the mass of the pilot is 67 kg

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