what is elastic collision in case of an elastic head on collision between two bodies derive an expression for the final velocities of the bodies in term of their masses and velocities before collision
Answers
Answer:
two particles are involved in an elastic collision, the velocity of the second particle after collision can be expressed as: v2f=2⋅m1(m2+m1)v1i+(m2−m1)(m2+m1)v2i v 2 f = 2 ⋅ m 1 ( m 2 + m 1 ) v 1 i + ( m 2 − m 1 ) ( m 2 + m 1 ) v 2 i .
Answer:
Conservation of Energy and Momentum
In an inelastic collision the total kinetic energy after the collision is not equal to the total kinetic energy before the collision.
LEARNING OBJECTIVES
Assess the conservation of total momentum in an inelastic collision
KEY TAKEAWAYS
Key Points
In an inelastic collision the total kinetic energy after the collision is not equal to the total kinetic energy before the collision.
If there are no net forces at work (collision takes place on a frictionless surface and there is negligible air resistance ), there must be conservation of total momentum for the two masses.
The variable θ is the angle between the velocity vector of the mass of interest and the x-axis in traditional Cartesian coordinate systems.
Key Terms
kinetic energy: The energy possessed by an object because of its motion, equal to one half the mass of the body times the square of its velocity.
momentum: (of a body in motion) the product of its mass and velocity.
At this point we will expand our discussion of inelastic collisions in one dimension to inelastic collisions in multiple dimensions. It is still true that the total kinetic energy after the collision is not equal to the total kinetic energy before the collision. While inelastic collisions may not conserve total kinetic energy, they do conserve total momentum.
We will consider an example problem in which one mass
(
m
1
)
slides over a frictionless surface into another initially stationary mass
(
m
2
)
. Air resistance will be neglected. The following things are known:
m
1
=
0.250
kg
,
m
2
=
0.400
kg
,
v
1
=
2.00
m
/
s
,
v
′
1
=
1.50
m
/
s
,
v
2
=
0
m
/
s
,
θ
′
1
=
45.0
∘
,
where
v
1
is the initial velocity of the first mass,
v
′
1
is the final velocity of the first mass,
v
2
is the initial velocity of the second mass, and
θ
′
1
is the angle between the velocity vector of the first mass and the x-axis.
The object is to calculate the magnitude and direction of the velocity of the second mass. After this, we will calculate whether this collision was inelastic or not.
Since there are no net forces at work (frictionless surface and negligible air resistance), there must be conservation of total momentum for the two masses. Momentum is equal to the product of mass and velocity. The initially stationary mass contributes no initial momentum. The components of velocities along the x-axis have the form
v
⋅
c
o
s
θ
, where θ is the angle between the velocity vector of the mass of interest and the x-axis.
Expressing these things mathematically:
m
1
v
1
=
m
1
v
′
1
⋅
c
o
s
(
θ
1
)
+
m
2
v
‘
2
⋅
c
o
s
(
θ
2
)
. (Eq. 2)
The components of velocities along the y-axis have the form v \cdot sin θ, where θ is the angle between the velocity vector of the mass of interest and the x-axis. By applying conservation of momentum in the y-direction we find:
0
=
m
1
v
′
1
⋅
s
i
n
(
θ
1
)
+
m
2
v
‘
2
⋅
s
i
n
(
θ
2
)
. (Eq. 3)
If we divide Eq. 3 by Eq. 2, we will find:
t
a
n
θ
2
=
v
′
1
⋅
s
i
n
θ
1
v
′
1
c
o
s
θ
θ
1
−
v
1
(Eq. 4)
Eq. 4 can then be solved to find
θ
2
approx. 312º.
Now let’ use Eq. 3 to solve for
v
′
2
. Re-arranging Eq. 3, we find:
v
′
2
=
−
m
1
v
′
1
⋅
s
i
n
θ
1
m
2
⋅
s
i
n
θ
2
.
After plugging in our known values, we find that
v
′
2
=
0.886
m
/
s
.
We can now calculate the initial and final kinetic energy of the system to see if it the same.
Initial Kinetic Energy =
1
2
m
1
⋅
v
2
1
+
1
2
m
2
⋅
v
2
2
=
0.5
J
.
Final Kinetic Energy =
1
2
m
1
⋅
v
′
1
2
+
1
2
m
2
⋅
v
′
2
2
≈
0.43
J
.
Since these values are not the same we know that it was an inelastic collision.
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