derive the equation for the velocity of the body before and after elastic collision
Answers
Answer:
Explanation:
Let us consider various types of two-object collisions. These collisions are the easiest to analyze, and they illustrate many of the physical principles involved in collisions. The conservation of momentum principle is very useful here, and it can be used whenever the net external force on a system is zero.
We start with the elastic collision of two objects moving along the same line—a one-dimensional problem. An elastic collision is one that also conserves internal kinetic energy. Internal kinetic energy is the sum of the kinetic energies of the objects in the system. Figure 1 illustrates an elastic collision in which internal kinetic energy and momentum are conserved.
Truly elastic collisions can only be achieved with subatomic particles, such as electrons striking nuclei. Macroscopic collisions can be very nearly, but not quite, elastic—some kinetic energy is always converted into other forms of energy such as heat transfer due to friction and sound. One macroscopic collision that is nearly elastic is that of two steel blocks on ice. Another nearly elastic collision is that between two carts with spring bumpers on an air track. Icy surfaces and air tracks are nearly frictionless, more readily allowing nearly elastic collisions on them.
ELASTIC COLLISION
An elastic collision is one that conserves internal kinetic energy.
INTERNAL KINETIC ENERGY
Internal kinetic energy is the sum of the kinetic energies of the objects in the system.
The system of interest contains a smaller mass m sub1 and a larger mass m sub2 moving on a frictionless surface. M sub 2 moves with velocity V sub 2 and momentum p sub 2 and m sub 1 moves behind m sub 2, with velocity V sub 1 and momentum p sub 1 toward the right direction. P 1 plus P 2 equals p total. The net force is zero. After collision m sub 1 moves toward the left with velocity V sub 1 while m sub 2 moves toward the right with velocity V sub 2 on the same frictionless surface. The momentum of m sub 1 becomes p 1 prime and m 2 becomes p 2 prime now. P 1 prime plus p 2 prime equals p total.
Figure 1. An elastic one-dimensional two-object collision. Momentum and internal kinetic energy are conserved.
Now, to solve problems involving one-dimensional elastic collisions between two objects we can use the equations for conservation of momentum and conservation of internal kinetic energy. First, the equation for conservation of momentum for two objects in a one-dimensional collision is
p1 + p2 = p′1 + p′2 (Fnet = 0)
or
m1v1 + m2v2 = m1v′1 + m2v′2 (Fnet = 0),
where the primes (′) indicate values after the collision. By definition, an elastic collision conserves internal kinetic energy, and so the sum of kinetic energies before the collision equals the sum after the collision. Thus,
1
2
m
1
v
1
2
+
1
2
m
2
v
2
2
=
1
2
m
1
v
'
1
2
+
1
2
m
2
v
'
2
2
(
two-object elastic collision
)
expresses the equation for conservation of internal kinetic energy in a one-dimensional collision.
EXAMPLE 1. CALCULATING VELOCITIES FOLLOWING AN ELASTIC COLLISION
Calculate the velocities of two objects following an elastic collision, given that m1 = 0.500 kg, m2 = 3.50 kg, v1 = 4.00 m/s, and v2 = 0.
Strategy and Concept
First, visualize what the initial conditions mean—a small object strikes a larger object that is initially at rest. This situation is slightly simpler than the situation shown in Figure 1 where both objects are initially moving. We are asked to find two unknowns (the final velocities v′1 and v′2). To find two unknowns, we must use two independent equations. Because this collision is elastic, we can use the above two equations. Both can be simplified by the fact that object 2 is initially at rest, and thus v2=0. Once we simplify these equations, we combine them algebraically to solve for the unknowns.
Solution
For this problem, note that v2=0 and use conservation of momentum. Thus,
p1 = p′1 + p′2 or m1v1=m1v′1+m2v′2.
Using conservation of internal kinetic energy and that v2=0,
1
2
m
1
v
1
2
=
1
2
m
1
v
'
1
2
+
1
2
m
2
v
'
2
2
Solving the first equation (momentum equation) for v′2, we obtain
v
'
2
=
m
1
m
2
(
v
1
−
v
'
1
)
.
Substituting this expression into the second equation (internal kinetic energy equation) eliminates the variable v′2, leaving only v′1 as an unknown (the algebra is left as an exercise for the reader). There are two solutions to any quadratic equation; in this example, they are
v′1 = 4 . 00 m/s and v′1=−3.00 m/s.