A collision of any two objects in physics is always either elastic or inelastic collision. Te elastic collision refers to a collision process where there is no loss in energy whereas the inelastic collision occurs with loss in energy of the system of the two objects that collide. In this article, we will study about elastic collision formula and its application.
Definition
An elastic collision happens when two bodies collide with each other, but, no loss occurs in the overall kinetic energy. In other words, it is an encounter amongst two bodies where the total kinetic energy of the two bodies is the same.
Moreover, it may be either one-dimensional or two-dimensional. Therefore, in the real world, perfectly elastic collision is not likely to happen as some conversion of energy, however small is bound to happen.
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Elastic Formula
An elastic collision is a collision where both the Kinetic Energy, KE, and momentum, p are conserved. In other words, it means that KE0 = KEf and po = pf. When we recall that KE = 1/2 mv2, we will write 1/2 m1(v1i)2 + 1/2 m2(vi)2 = 1/2 m1(v1f)2 + 1/2 m2 (v2f)2. Thus, we see that the final total KE of the bodies is similar to the initial KE of these two bodies. Moreover, as p = linear momentum = mv, then we will write m1v1i + m2v2i = m1v1f + m2v2f.
[A] m1v1i + m2v2i = m1v1f + m2v2f
[B] 1/2 m1(v1i)2Â + 1/2 m2(vi)2Â = 1/2 m1(v1f)2Â + 1/2 m2Â (v2f)2
Over here:
KE is the kinetic energy
p refers to the momentum
m is the mass, kg
mi refers to the mass of the 1st object
and m2 is the mass of the 2nd object
v is the velocity, m/s
v1Â refers to the velocity of the 1st object
and v2Â is the velocity of the 2nd object
vi refers to the initial velocity
vf is the final velocity
Solved Examples on Elastic Formula
Question- A green ball having a mass of 0.2 kg hits a yellow ball having a mass of 0.25 kg in an elastic collision, and the green ball halts. After that, the velocity of the green ball is 5 m/s and the yellow ball was at rest. Calculate the final velocity of the yellow ball.
Answer- Looking at the question, we see we have the values of the mass of the 1st ball, m1 = 0.2 kg and the the mass of the 2nd ball, m2 = 0.25kg. Furthermore, we have the initial velocity of the 1st ball, v1i = 5 m/s and the initial velocity of the 2nd ball, v2i = 0. Finally, we have the final velocity of the 1st ball, (v1f) = 0.
m1v1i + m2v2i = m1v1f + m2v2f
(0.2 kg)(5 m/s) + (0.25 kg)(0 m/s) = (0.2 kg)(0) + (0.25 kg)(v2f)
1.0 kg.m/s + 0 = 0 + (0.25 kg)(v2f)
1.0 kg.m/s = (0.25 kg)(v2f)
(1.0 kg.m/s) / 0.25 kg = (v2f)
4 m/s = (v2f)
Question- Find out the final velocity of the yellow ball using the equation for conservation of kinetic energy in an elastic collision.
Answer- We have the mass of the 1st ball, m1 = 0.2 kg and the mass of the 2nd ball, m2 = 0.20kg. Furthermore, the initial velocity of the 1st ball is given as v1i = 5 m/s and the initial velocity of the 2nd ball, v2i = 0. Finally, the final velocity of the 1st ball, (v1f) = 0.
1/2 m1(v1i)2Â + 1/2 m2(vi)2Â = 1/2 m1(v1f)2Â + 1/2 m2Â (v2f)2
1/2 (0.2 kg)(5m/s)2Â + 1/2 (0.2 kg)(0) = 1/2 (0.2 kg)(0) + 1/2 (0.2 kg)(v2f)2
1/2 (0.2 kg)(5m/s)2Â = 1/2 (0.2 kg)(v2f)2
(5m/s)2Â = (v2f)2
25 m2/s2Â = (v2f)2
v2f = √25 m2/s2
v2f = 5 m/s
Typo Error>
Speed of Light, C = 299,792,458 m/s in vacuum
So U s/b C = 3 x 10^8 m/s
Not that C = 3 x 108 m/s
to imply C = 324 m/s
A bullet is faster than 324m/s
I have realy intrested to to this topic
m=f/a correct this
Interesting studies
It is already correct f= ma by second newton formula…