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san4es73 [151]
4 years ago
11

A 50g marble is moving at 2m/s when it strikes a 20g marble at rest. Immediately after the collision, the 50g ball is moving at

1m/s. Is this an elastic collision
Physics
1 answer:
Alika [10]4 years ago
8 0

Answer:

Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.

Explanation:

Given;

mass of first marble, m₁ = 50g = 0.05 kg

initial velocity of the first marble, u₁ = 2 m/s

mass of second marble, m₂ = 20 g = 0.02 kg

initial velocity of the second marble, u₂ = 0

final velocity of the first marble, v₁ = 1 m/s

Let the final velocity of the second marble, = v₂

Determine the final velocity of the second marble by applying principle of momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

0.05 x 2 + 0.02 x 0 = 0.05 x 1 + 0.02v₂

0.1 = 0.05 + 0.02v₂

0.02v₂ = 0.1 - 0.05

0.02v₂ = 0.05

v₂ = 0.05 / 0.02

v₂ = 2.5 m/s

During inelastic collision both objects will move at the same velocity after collision.

During elastic collision both objects will move at different velocities after collision.

Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.

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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,
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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,
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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
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A
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​

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

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​
v
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​
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A
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