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Virty [35]
2 years ago
12

Write down the conservation of momentum?​

Physics
1 answer:
otez555 [7]2 years ago
5 0
Law of conservation of momentum states that when two objects collide with each other , the sum of their linear momentum always remains same or we can say conserved and is not effected by any action, reaction only in case is no external unbalanced force is applied on the bodies.
Let,
m
A
​
= Mass of ball A
m
B
​
= Mass of ball B
u
A
​
= initial velocity of ball A
u
B
​
= initial velocity of ball B
v
A
​
= Velocity after the collision of ball A
v
B
​
= Velocity after the collision of ball B
F
ab
​
= Force exerted by A on B
F
ba
​
= Force exerted by B on A
Now,
Change in the momentum of A= momentum of A after the collision - the momentum of A before the collision
= m
A
​
v
A
​
−m
A
​
u
A
​

Rate of change of momentum A= Change in momentum of A/ time taken
=
t
m
A
​
v
A
​
−m
A
​
u
A
​

​

Force exerted by B on A (F
ba
​
);
F
ba
​
=
t
m
A
​
v
A
​
−m
A
​
u
A
​

​
........ [i]
In the same way,
Rate of change of momentum of B=
t
m
b
​
v
B
​
−m
B
​
u
B
​

​

Force exerted by A on B (F
ab
​
)=
F
ab
​
=
t
m
B
​
v
B
​
−m
B
​
u
B
​

​
.......... [ii]
Newton's third law of motion states that every action has an equal and opposite reaction, then,
F
a
​
b=−F
b
​
a [ ' -- ' sign is used to indicate that 1 object is moving in opposite direction after collision]

Using [i] and [ii] , we have
t
m
B
​
v
B
​
−m
B
​
u
B
​

​
=−
t
m
A
​
v
A
​
−m
A
​
u
A
​

​

m
B
​
v
B
​
−m
B
​
u
B
​
=−m
A
​
v
A
​
+m
A
​
u
A
​

Finally we get,
m
B
​
v
B
​
+m
A
​
v
A
​
=m
B
​
u
B
​
+m
A
​
u
A
​

This is the derivation of conservation of linear momentum.
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A ball is thrown horizontally from the top of a 55 m building and lands 150 m from the base of the building. Ignore air resistan
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Answer:

a) t =3.349 s

b) V_x,i = 44.8 m/s

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Explanation:

Given:

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Find:

a) How long is the rock in the air in seconds.  

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

                                 y(f) = y(0) + V_y,i*t + 0.5*g*t^2

- V_y,i = 0 (horizontal throw)

                                 55 = 0 + 0 + 0.5*(9.81)*t^2

                                 t = sqrt ( 55 * 2 / 9.81 )

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- Use the second equation of motion in x direction:

                                 x(f) = x(0) + V_x,i*t

                                 150 = 0 + V_x,i*3.349

                                  V_x,i = 150 / 3.349 = 44.8 m/s

- Use the first equation of motion in y direction:

                                 V_y,f = V_y,i + g*t

                                 V_y,f = 0 + 9.81*3.349

                                 V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

                                 V^2 = V_y,f^2 + V_x,i^2

                                 V = sqrt (32.85^2 + 44.8^2)

                                 V = 55.55 m/s

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We can write therefore the energy difference between adjacent levels as

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We see that this difference decreases as the level number (n) increases. For example, the difference between the levels n=1 and n=2 is

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While the difference between the levels n=2 and n=3 is

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And so on.

So, the energy difference between adjacent levels decreases as the quantum number increases.

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