A collision between two particles is said to occur if they physically strike against each other or if the path of the motion of one is influenced by the other. In physics, the term collision does not necessarily mean that a particle actually strikes. In fact, two particles may not even touch each other and yet they are said to collide if one particle influences the motion of the other. When two bodies collide, each body exerts a force on the other. The two forces are exerted simultaneously for an equal but short interval of time. According to Newton’s third law of motion, each body exerts an equal and opposite force on the other at each instant of collision. During a collision, the two fundamental conservation laws namely, the law of conservation of momentum and that of energy are obeyed and these laws can be used to determine the velocities of the bodies after collision. Collisions are divided into two types : (i) elastic collision and (ii) inelastic collision Elastic collision If the kinetic energy of the system is conserved during a collision, it is called an elastic collision. (i.e) The total kinetic energy before collision and after collision remains unchanged. The collision between subatomic particles is generally elastic. The collision between two steel or glass balls is nearly elastic. In elastic collision, the linear momentum and kinetic energy of the system are conserved. Elastic collision in one dimension. If the two bodies after collision move in a straight line, the collision is said to be of one dimension.Consider two bodies A and B of masses m1 and m2 moving along the same straight line in the same direction with velocities u1 and u2 respectively as shown in Fig. below. Let us assume that u1 is greater than u2. The bodies A and B suffer a head on collision when they strike and continue to move along the same straight line with velocities v1 and v2 respectively. From the law of conservation of linear momentum, Total momentum before collision = Total momentum after collision m1u1 + m2u2 = m1v1 + m2v2 …(1) Since the kinetic energy of the bodies is also conserved during the collision Total kinetic energy before collision = Total kinetic energy after collision Equation (5) shows that in an elastic one-dimensional collision, the relative velocity with which the two bodies approach each other before collision is equal to the relative velocity with which they recede from each other after collision.
Special cases Case ( i) : If the masses of colliding bodies are equal, i.e. m1 = m2 v1 = u2 and v2 = u1 …(9) After head on elastic collision, the velocities of the colliding bodies are mutually interchanged. Case (ii) : If the particle B is initially at rest, (i.e) u2 = 0 then v1 = (ma – mb)/(ma+mb)ua …(10) and v2 =(2ma)/(ma+mb)U1 …(11) Inelastic collision During a collision between two bodies if there is a loss of kinetic energy, then the collision is said to be an inelastic collision. Since there is always some loss of kinetic energy in any collision, collisions are generally inelastic. In inelastic collision, the linear momentum is conserved but the energy is not conserved. If two bodies stick together, after colliding, the collision is perfectly inelastic but it is a special case of inelastic collision called plastic collision. (eg) a bullet striking a block of wood and being embedded in it. The loss of kinetic energy usually results in the form of heat or sound energy. Let us consider a simple situation in which the inelastic head on collision between two bodies of masses mA and mB takes place. Let the colliding bodies be initially move with velocities u1 and u2. After collision both bodies stick together and moves with common velocity v. Total momentum of the system before collision = mAu1 + mBu2 Total momentum of the system after collision = mass of the composite body × common velocity = (mA+ mB ) v By law of conservation of momentum
It is clear from the above equation that in a perfectly inelastic collision, the kinetic energy after impact is less than the kinetic energy before impact. The loss in kinetic energy may appear as heat energy.
By the end of this section, you will be able to:
In the previous two sections, we considered only one-dimensional collisions; during such collisions, the incoming and outgoing velocities are all along the same line. But what about collisions, such as those between billiard balls, in which objects scatter to the side? These are two-dimensional collisions, and we shall see that their study is an extension of the one-dimensional analysis already presented. The approach taken (similar to the approach in discussing two-dimensional kinematics and dynamics) is to choose a convenient coordinate system and resolve the motion into components along perpendicular axes. Resolving the motion yields a pair of one-dimensional problems to be solved simultaneously. One complication arising in two-dimensional collisions is that the objects might rotate before or after their collision. For example, if two ice skaters hook arms as they pass by one another, they will spin in circles. We will not consider such rotation until later, and so for now we arrange things so that no rotation is possible. To avoid rotation, we consider only the scattering of point masses—that is, structureless particles that cannot rotate or spin. We start by assuming that Fnet=0, so that momentum p is conserved. The simplest collision is one in which one of the particles is initially at rest. (See Figure 1.) The best choice for a coordinate system is one with an axis parallel to the velocity of the incoming particle, as shown in Figure 1. Because momentum is conserved, the components of momentum along the x– and y-axes (px and py) will also be conserved, but with the chosen coordinate system, py is initially zero and px is the momentum of the incoming particle. Both facts simplify the analysis. (Even with the simplifying assumptions of point masses, one particle initially at rest, and a convenient coordinate system, we still gain new insights into nature from the analysis of two-dimensional collisions.) Along the x-axis, the equation for conservation of momentum is p1x + p2x = p′1x + p′2x. Where the subscripts denote the particles and axes and the primes denote the situation after the collision. In terms of masses and velocities, this equation is m1v1x + m2v2x = m2v2x = m1v′1x + m2v′2x. But because particle 2 is initially at rest, this equation becomes m1v1x = m1v′1x + m2v′2x. The components of the velocities along the x-axis have the form v cos θ. Because particle 1 initially moves along the x-axis, we find v1x = v1. Conservation of momentum along the x-axis gives the following equation: m1v1 = m1v′1 cos θ1 + m2v′2 cos θ2, where θ1 and θ2 are as shown in Figure 1.
m1v1 = m1v′1 cos θ1 + m2v′2 cos θ2 Along the y-axis, the equation for conservation of momentum is p1y + p2y = p′1y + p′2y or m1v1y + m2v2y = m1v′1y + m2v′2y. But v1y is zero, because particle 1 initially moves along the x-axis. Because particle 2 is initially at rest, v2y is also zero. The equation for conservation of momentum along the y-axis becomes 0 = m1v′1y + m2v′2y. The components of the velocities along the y-axis have the form v sin θ. Thus, conservation of momentum along the y-axis gives the following equation: 0 = m1v′1 sin θ1 + m2v′2 sin θ2.
0 = m1v′1 sin θ1 + m2v′2 sin θ2 The equations of conservation of momentum along the x-axis and y-axis are very useful in analyzing two-dimensional collisions of particles, where one is originally stationary (a common laboratory situation). But two equations can only be used to find two unknowns, and so other data may be necessary when collision experiments are used to explore nature at the subatomic level.
Suppose the following experiment is performed. A 0.250-kg object (m1) is slid on a frictionless surface into a dark room, where it strikes an initially stationary object with mass of 0.400 kg (m2). The 0.250-kg object emerges from the room at an angle of 45.0º with its incoming direction. The speed of the 0.250-kg object is originally 2.00 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity (v′2 and θ2) of the 0.400-kg object after the collision. StrategyMomentum is conserved because the surface is frictionless. The coordinate system shown in Figure 2 is one in which m2 is originally at rest and the initial velocity is parallel to the x-axis, so that conservation of momentum along the x– and y-axes is applicable. Everything is known in these equations except v′2 and θ2, which are precisely the quantities we wish to find. We can find two unknowns because we have two independent equations: the equations describing the conservation of momentum in the x– and y-directions. SolutionSolving m1v1 = m1v′1 cos θ1 + m2v′2 cos θ2 for v′2 cos θ2 and 0 = m1v′1 sin θ1 + m2v′2 sin θ2 for v′2 sin θ2 and taking the ratio yields an equation (in which θ2 is the only unknown quantity. Applying the identity [latex]\displaystyle\left(\tan\theta=\frac{\sin\theta}{\cos\theta}\right)\\[/latex], we obtain: [latex]\displaystyle\tan\theta_{2}=\frac{v′_{1}\sin\theta_1}{v′_{1}\cos\theta_1-v_1}\\[/latex] Entering known values into the previous equation gives [latex]\tan{\theta }_{2}=\frac{\left(1\text{.}\text{50}\text{m/s}\right)\left(0\text{.}\text{7071}\right)}{\left(1\text{.}\text{50}\text{m/s}\right)\left(0\text{.}\text{7071}\right)-2\text{.}\text{00}\text{m/s}}=-1\text{.}\text{129}\\[/latex]. Thus, θ2 = tan−1 (−1.129) = 311.5º ≈ 312º. Angles are defined as positive in the counter clockwise direction, so this angle indicates that m2 is scattered to the right in Figure 2, as expected (this angle is in the fourth quadrant). Either equation for the x– or y-axis can now be used to solve for v′2, but the latter equation is easiest because it has fewer terms. [latex]\displaystyle{v}′_{2}=\frac{m_1}{m_2}v′_{1}\frac{\sin\theta_{1}}{\sin\theta_{1}}\\[/latex] Entering known values into this equation gives [latex]\displaystyle{v}′_{2}=-\left(\frac{0.250\text{ kg}}{0.400\text{ kg}}\right)\left(1.50\text{ m/s}\right)\left(\frac{0.7071}{-0.7485}\right)\\[/latex] Thus, v′2 = 0.886 m/s. DiscussionIt is instructive to calculate the internal kinetic energy of this two-object system before and after the collision. (This calculation is left as an end-of-chapter problem.) If you do this calculation, you will find that the internal kinetic energy is less after the collision, and so the collision is inelastic. This type of result makes a physicist want to explore the system further. Elastic Collisions of Two Objects with Equal MassSome interesting situations arise when the two colliding objects have equal mass and the collision is elastic. This situation is nearly the case with colliding billiard balls, and precisely the case with some subatomic particle collisions. We can thus get a mental image of a collision of subatomic particles by thinking about billiards (or pool). (Refer to Figure 1 for masses and angles.) First, an elastic collision conserves internal kinetic energy. Again, let us assume object 2 (m2) is initially at rest. Then, the internal kinetic energy before and after the collision of two objects that have equal masses is [latex]\frac{1}{2}{mv_1}^2=\frac{1}{2}{mv′_{1}}^2+\frac{1}{2}{mv′_{2}}^{2}\\[/latex]. Because the masses are equal, m1 = m2 = m. Algebraic manipulation (left to the reader) of conservation of momentum in the x– and y-directions can show that [latex]\frac{1}{2}mv_1^2=\frac{1}{2}{mv′_{1}}^2+\frac{1}{2}{mv′_{2}}^{2}+mv′_{1}v′_{2}\cos\left(\theta_{1}-\theta_{2}\right)\\[/latex]. (Remember that θ2 is negative here.) The two preceding equations can both be true only if mv′1v′2 cos(θ1 − θ2) = 0. There are three ways that this term can be zero. They are
All three of these ways are familiar occurrences in billiards and pool, although most of us try to avoid the second. If you play enough pool, you will notice that the angle between the balls is very close to 90º after the collision, although it will vary from this value if a great deal of spin is placed on the ball. (Large spin carries in extra energy and a quantity called angular momentum, which must also be conserved.) The assumption that the scattering of billiard balls is elastic is reasonable based on the correctness of the three results it produces. This assumption also implies that, to a good approximation, momentum is conserved for the two-ball system in billiards and pool. The problems below explore these and other characteristics of two-dimensional collisions.
Two-dimensional collision experiments have revealed much of what we know about subatomic particles, as we shall see in Medical Applications of Nuclear Physics and Particle Physics. Ernest Rutherford, for example, discovered the nature of the atomic nucleus from such experiments. Section Summary
[latex]\frac{1}{2}{mv_{1}}^{2}=\frac{1}{2}{mv′_{1}}^{2}+\frac{1}{2}{mv′_{2}}^{2}+{\text{mv}′}_{1}{v′}_{2}\text{cos}\left({\theta }_{1}-{\theta }_{2}\right)\\[/latex].
Problems & Exercises
point masses: structureless particles with no rotation or spin
1. (a) 3.00 m/s, 60º below x-axis; (b) Find speed of first puck after collision: [latex]\displaystyle{0}=m{v′ }_{1}\sin30^{\circ}-m{v′ }_{2}\sin60^{\circ}\Rightarrow {v′ }_{1}^{}={v′ }_{2}^{}\frac{\sin60^{\circ}}{\sin30^{\circ}}=\text{5.196 m/s}\\[/latex] Verify that ratio of initial to final KE equals one: [latex]\begin{cases}\text{KE}=\frac{1}{2}{mv}_{1}^{2}=18m\text{ J}\\ \text{KE}=\frac{1}{2}{mv′_{1}}^{2}+\frac{1}{2}{mv′_{2}}^{2}=18m\text{ J}\end{cases}\\[/latex] [latex]\frac{\text{KE}}{\text{KE}}′=1.00\\[/latex] 3. (a) −2.26m/s; (b) 7.63 × 103 J; (c) The ground will exert a normal force to oppose recoil of the cannon in the vertical direction. The momentum in the vertical direction is transferred to the earth. The energy is transferred into the ground, making a dent where the cannon is. After long barrages, cannon have erratic aim because the ground is full of divots. 5. (a) 5.36 × 105 m/s at −29.5º; (b) 7.52 × 10−13 J 7. We are given that [latex]{m}_{1}={m}_{2}\equiv m\\[/latex]. The given equations then become: [latex]{v}_{1}={v}_{1}\text{cos}{\theta }_{1}+{v}_{2}\text{cos}{\theta }_{2}\\[/latex] [latex]0={v′}_{1}^{}\text{sin}{\theta }_{1}+{v′}_{2}^{}\text{sin}{\theta }_{2}\\[/latex]. Square each equation to get [latex]\begin{array}{lll}{{v}_{1}}^{2}& =& {{v′ }_{1}}^{2}{\text{cos}}^{2}{\theta }_{1}+{{v′ }_{2}}^{2}{\text{cos}}^{2}{\theta }_{2}+2{v′ }_{1}{v′}_{2}\text{cos}{\theta }_{1}\text{cos}{\theta }_{2}\\ 0& =& {{v′ }_{1}}^{2}{\text{sin}}^{2}{\theta }_{1}+{{v′ }_{2}}^{2}{\text{sin}}^{2}{\theta }_{2}+2{v′ }_{1}{v′ }_{2}\text{sin}{\theta }_{1}\text{sin}{\theta }_{2}\text{.}\end{array}\\[/latex] Add these two equations and simplify: [latex]\begin{array}{lll}{{v}_{1}}^{2}& =& {{v′ }_{1}}^{2}+{{v′ }_{2}}^{2}+2{{v′ }_{1}}^{}{{v′ }_{2}}^{}\left(\text{cos}{\theta }_{1}\text{cos}{\theta }_{2}+\text{sin}{\theta }_{1}\text{sin}{\theta }_{2}\right)\\ & =& {{v′ }_{1}}^{2}+{{v′ }_{2}}^{2}+2{v′ }_{1}{v′ }_{2}\left[\frac{1}{2}\text{cos}\left({\theta }_{1}-{\theta }_{2}\right)+\frac{1}{2}\text{cos}\left({\theta }_{1}+{\theta }_{2}\right)+\frac{1}{2}\text{cos}\left({\theta }_{1}-{\theta }_{2}\right)-\frac{1}{2}\text{cos}\left({\theta }_{1}+{\theta }_{2}\right)\right]\\ & =& {{v′ }_{1}}^{2}+{{v′ }_{2}}^{2}+2{v′ }_{1}{v′ }_{2}\text{cos}\left({\theta }_{1}-{\theta }_{2}\right).\end{array}\\[/latex] Multiply the entire equation by [latex]\frac{1}{2}m\\[/latex] to recover the kinetic energy: [latex]\frac{1}{2}{{\mathit{\text{mv}}}_{1}}^{2}=\frac{1}{2}m{{v′ }_{1}}^{2}+\frac{1}{2}m{{v′ }_{2}}^{2}+m{v′ }_{1}{v′ }_{2}\text{cos}\left({\theta }_{1}-{\theta }_{2}\right)\\[/latex] as discussed in the text. |