Physics Topics can help us understand the behavior of the natural world around us.
What are the Three Conservation Equations?
Let a particle of mass m1 moving with a velocity u1 along AO, collide at the point O with another particle of mass m2 moving with velocity u2 along BO [Fig.], The point O is chosen as the origin, the direction of AO as the x-axis,
and a perpendicular direction on the plane of AO and BO as the y -axis.
Before the collision, it is evident from Fig. that,
x – component of momentum of the 1st particle = m1u1,
y -component of momentum of the 1st particle = 0,
x -component of momentum of the 2nd particle = -m2u2 cosθ,
y-component of momentum of the 2nd particle = m2u2 sinθ.
So, there is no component of momentum perpendicular to the xy-plane, i.e., along the z -direction. We know that, in both elastic and inelastic collisions, the total momentums of the particles remain conserved. Thus, even after the collision between the two particles, there will be no z -component of momentum. This means that, every two-particle collision, in general is two-dimensional, i.e., such a collision is always confined in a plane.
In collision experiments, usually a beam of incident particles hits a stationary target. After collision, the first particle is observed by placing a detector D in a definite orientation, say, at right angles with the direction of the incident beam [Fig.].
In the figure,
m1, m2 : masses of the particles;
u1 = velocity of m1 before collision;
u2 = 0 = velocity of m2 before collision;
v1 = velocity of m1 after collision deviated by 90°;
v2 = velocity of m2 after collision;
θ = angle made by v2 with the direction of the incident beam.
Now, we may apply the principle of conservation of momentum, with the choice of the x and y-axes as shown in the figure.
x-direction: m1u1 + 0 = 0 + m2v2cosθ
or, m2v2cosθ = m1u1 ……….. (1)
y-direction: 0 + 0 = m1v1 – m2v2sinθ
or, m2vsinθ = m1v1 ……… (2)
Squaring and adding equations (1) and (2),
m2v2 = m1\(\sqrt{u_1^2+v_1^2}\) or, v2 = \(\frac{m_1}{m_2} \sqrt{u_1^2+v_1^2}\) ………. (3)
Again, dividing equation (2) by equation (1),
tanθ = \(\frac{v_1}{u_1}\), θ = tan-1\(\frac{v_1}{u_1}\) …….(4)
Usually, m1, m2 and are known prior to the experiment, and v1 is measured by the detector D. So the magnitude and direction of the velocity (v2) of the second particle after collision can be calculated using equations (3) and (4).
General analysis of two-dimensional collisions: We have already seen that the motion of two particles, both before and after they collide with each other, is confined in a plane. Let that plane be chosen as the xy -plane. We assume that, the magnitudes and the directions of the velocities (i.e., of the momenta) of the two particles, before the collision, are known beforehand. Then, four unknown quantities, associated with the collision, are to be solved for. They are the magnitudes and the directions of the velocities of the two particles after the collision. If the collision is elastic, we can construct at most three equations from the conservation principles:
- conservation of momentum along the x-axis,
- conservation of momentum along the y-axis,
- conservation of kinetic energy.
These three equations are not sufficient to solve for four unknown quantities. If the collision is inelastic, the kinetic energy is not conserved. Then we have only two equations at hand. So, for a complete analysis of a two-particle collision, we should have some information on the particles after the collision. For example, in the collision experiment described above, we had placed a detector in such a way—at right angles to the initial direction of motion of one particle—that the final direction of the velocity of that particle was already assigned.
Moreover, the detector could measure the magnitude of its velocity. So, we had to solve for only two unknown quantities—the magnitude and direction of the velocity of the second particle. So, only two equations were sufficient— equations obtained from the momentum conservation along the x- and the y-axes.
Even the kinetic energy equation was not necessary; so the treatment was applicable to elastic as well as inelastic collisions. One other point is to be noted. In general, both the particles may be in motion before the collision [Fig.(a)]. But the collision may be observed by assuming one of the particles to be at rest. In Fig.(b), the particle 2 is at rest; then the velocity of particle 1 is actually its relative velocity with respect to particle 2. This means that the frame attached to particle 2 has been taken as the frame of reference. So, there is no loss of generality if we describe a two-particle collision as a moving particle colliding with a stationary target.
Numerical Examples
Example 1.
Two particles of masses m1 and m2, moving with velocities u1 and u2, respectively and making an angle θ between them, collide with each other. After collision, the 1st particle travels In the initial direc-tion of motion of the 2nd, and vice-versa. Find the velocities of the two particles after collision. Under what condition, would this collision be elastic?
Solution:
Suppose v1, v2 are the velocities of the two particles, respectively, after collision. The particles before and after collision move as shown in Fig. It also shows the chosen directions of the x and the y-axes.
For momentum conservation along the x -axis, we get,
Here, K1 ≠ K2; so the collision is inelastic, in general. As a special case, it would be an elastic collision if K1 = K2. It is possible only when m1 = m2, i.e., the two particles are of equal masses.
Example 2.
A bomb explodes and splits up into three fragments. Two fragments, each of mass 200 g, move away from each other making an angle 120°, at a speed of 100 m ᐧ s-1. Find the direction and velocity of the third fragment whose mass is 500 g. Also find out the energy released in explosion.
Solution:
Fig. shows the velocity of fragments A and B along OA and OB. A and B have equal mass and speed. From the law of conservation of linear momentum, the third piece must move along OD, in the direction opposite to the resultant of OA and OB.
If the velocity of the third piece is v, then taking the components along the line CD in CGS system,
500v = 200 × 104cos60° + 200 × 104cos60°
or, v = \(\frac{200 \times 10^4}{500}\) = 4 × 103 cm ᐧ s-1
Hence, the velocity of the third fragment is 40 m ᐧ s-1. It moves so as to make an angle 120° with each of OA and OB. The energy released due to explosion, is the kinetic energy of the three fragments.
∴ Energy released
= \(\frac{1}{2}\) × 200 × (104)2 + \(\frac{1}{2}\) × 200 × (104)2 + \(\frac{1}{2}\) × 500 × (4 × 103)2
= 2400 × 107 ergs = 2400 J.
Example 3.
A spaceship while flying in space, splits up into three equal parts, due to an explosion. One fragment keeps moving in the same direction; the other two fly off at 60° to the original direction, on either side. If the energy released due to the explosion is twice the kinetic energy of the spaceship, find the kinetic energy of each of the fragments.
Solution:
Let the mass of the spaceship be 3m and initial speed be u.
Hence, mass of each frag-ment = m. Let their velocities be v1, v2 and v3 after the explosion [Fig.].
Applying the conservation of momentum law along the direction perpendicular to the original direction of motion,
0 = mv2sin60° – mv3sin60°
or, mv2sin60° = mv3sin60°
or, v2 = v3 = v (say)
Applying the law of conservation of linear momentum along the original direction of motion,
3mu = mv2 + 2mv cos60°
or, 3u = v1 + v ……… (1)
Kinetic energy of the spaceship before explosion
E = \(\frac{1}{2}\) × 3mu2
Energy released during the explosion,
Er = \(\frac{1}{2} m\left(v_1^2+2 v^2\right)\)+ 2v2) – \(\frac{1}{2} \cdot 3 m u^2\)
According to the problem,