Contents
Physics Topics are also essential for space exploration, allowing scientists to study phenomena such as gravitational waves and cosmic rays.
Is Momentum Kinetic Energy and Total Energy Conserved in an Inelastic Collision?
Total momentum of a system of interacting bodies remains constant in the absence of an external force. But generally, the total kinetic energy of the system is not conserved. In most collisions, a fraction of the kinetic energy transforms into heat and sound.
If the total momentum and the total kinetic energy of a system are conserved, the collision Is termed as an elastic collision. Collision of two billiard balls on a smooth board, is almost elastic in nature. Collisions in atoms and molecules, or interactions involving protons, electrons, neutrons, etc., are assumed to be elastic.
On the other hand, if the total momentum is conserved, but the total kinetic energy is not, it is an inelastic collision. Most practical collisions are inelastic.
One-dimensional Elastic Collision between Two Particles
Suppose m1 and m2 are the masses of two particles which are moving with velocities \(\overrightarrow{u_1}\) and \(\overrightarrow{u_2}\) respectively (u1 > u2) in the same direction along a straight line. They collide elastically, and after collision, move along the same direction with velocities \(\overrightarrow{v_1}\) and \(\overrightarrow{v_2}\) respectively [Fig.].
Since one-dimensional motion is considered here vector notation can be dropped. Only components can be used with sign to indicate direction.
From the law of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
or, m1(u1 – v1) = m2(v2 – u2) ………. (1)
In an elastic collision, the total kinetic energy of the particles will also be conserved.
Hence, kinetic energy before collision = kinetic energy after collision
Similarly, taking the value of v1 from equation (3) and substituting it in equation (1),
v2 = \(\frac{m_2-m_1}{m_1+m_2} u_2\) + \(\frac{2 m_1}{m_1+m_2} u_1\) ………. (5)
It is important to note that, equations (4) and (5) are symmetrical against interchange of the first and the second particles i.e. if the subscripts 1 and 2 are interchanged in Fig., same equations (4) and (5) will be obtained.
Special Cases:
i) Particles are of equal mass:
In this case m1 = m2; hence from equations (4) and (5), v1 = u2 and v2 = u1.
Particles exchange their velocities after collision.
ii) Particles are of equal mass, and the second particle is initially at rest:
Here, m1 = m2 and u2 = 0. Hence v1 = 0 and v2 = u1, from equations (4) and (5) . The first particle comes to rest and the second particle gains the velocity of the first, after collision. Such events are frequent in games like billiard.
iii) Particles are of unequal masses and the second particle is initially at rest.
Here, m1 ≠ m2 and u2 = 0. Equations (4) and (5) thus change as,
v1 = \(\frac{m_1-m_2}{m_1+m_2} u_1\) and v2 = \(\frac{2 m_1}{m_1+m_2} u_1\)
If m1 and m2 have different values, the first particle will not stop after collision. Hence, when a striker hits a stationary coin in a game of carrom, both the striker and the coin move after collision.
iv) First particle is much heavier than the second particle and the second particle is initially at rest.
In this case, m1 \(\gg\) m2 and u2 = 0.
Hence, m1 – m2 \(\simeq\) < m1 + m2 \(\simeq\) m1
∴ v1 \(\simeq\) u1 and v2 \(\simeq\) 2u1
Therefore, after collision, the velocity of the first (massive) particle practically remains unchanged; but the second particle gains a velocity equal to almost twice the initial velocity of the first particle. In a collision, the velocity acquired by a body cannot be greater than twice the velocity of the collider.
v) Second particle is much heavier than the first and is initially at rest.
Here, m2 \(\gg\) m1 and u2 = 0. Hence m1 – m2 ≈ m2
Values of v1 and v2 are thus v2 \(\simeq\) and v2 = 0.
Hence, after the collision, the massive body will continue to be at rest; and the collider will recoil with the same magnitude of velocity. A collision between a tennis ball and the earth’s surface is of this type.
Numerical Examples
Example 1.
Two particles of equal mass moving towards each other with velocities 20m ᐧ s-1 and 30 m ᐧ s-1 collide. If the collision is elastic, find their velocities after the collision.
Solution:
The particles are moving towards each other. If the velocity (u1) of one is 20 m ᐧ s-1, then the velocity (u2) of the other particle is -30 m ᐧ s-1.
After collision, suppose the velocities are v1 and v2 respectively
From the law of conservation of linear momentum,
m × 20 – m × 30 = mv1 + mv2
or, v1 + v2 = -10 …… (1)
For elastic collision,
v2 – v1 = u1 – u2
or, v2 – v1 = 20 – (-30) = 50 ……… (2)
Solving equation (1) and (2),
v1 = -30 m ᐧ s-1 and v2 = 20 m ᐧ s-1.
Example 2.
Three balls A, B, C of masses m1, m2 and m3 respectively are kept at rest along a straight line. Now A moving in that straight line with velocity u1 strikes B and then B moving with velocity u2 strikes C. As a result velocity of C becomes u3. If the collisions are elastic, show that u3 \(\simeq\)4u1, when m1 \(\gg\) m2 and m2 \(\gg\) m3. In case A hits C directly, will the velocity of C be higher or lower?
Solution:
Suppose A acquires a velocity v1 after collision with B. From the law of conservation of linear momentum, for collision between A and B,
m1u1 = m1v1 + m2u2
or, m1(u1 – v1) = m2u2 ……… (1)
From the law of conservation of kinetic energy for elastic collision,
Similarly, for the collision of B with C, u3 ≈ 2u2. Thus u3 \(\simeq\) 4u1
But, for the direct collision of A with C, the velocity gained by C, u3‘ ≈ 2u1.
∴ u3‘ is less than u3.
Hence, in case of direct collision between A and C, the speed of C will be less.