Lesson Explainer: Collisions and Conservation of Momentum | Nagwa Lesson Explainer: Collisions and Conservation of Momentum | Nagwa

Lesson Explainer: Collisions and Conservation of Momentum Mathematics

In this explainer, we will learn how to apply the law of conservation of momentum to study collisions in one dimension and differentiate between elastic and inelastic collisions.

Let us first recall the relationship between the impulse produced by the action of a force and the change in momentum.

Property: Impulse and Change of Momentum

For a constant-mass body, the impulse produced by the action of a force over an interval of time equals the change in momentum of the body: 𝐽=𝐹𝑡=Δ𝑝.d

Consider two particles, 1 and 2, with respective momenta 𝑝i and 𝑝i, as shown in the following diagram. During the time of collision, particle 1 exerts force 𝐹 on particle 2, and particle 2 exerts force 𝐹 on particle 1. These two forces result from the interaction between particle 1 and particle 2. Newton’s third law of motion tells us that these forces are equal in magnitude but opposite in direction. Hence, we have 𝐹=𝐹.

Particle 1 experiences a change in momentum caused by the action of the force exerted by particle 2 during the collision (between 𝑡 and 𝑡) given by Δ𝑝=𝑝𝑝=𝐽=𝐹𝑡.fid

Similarly, the change in momentum of particle 2 is given by Δ𝑝=𝑝𝑝=𝐽=𝐹𝑡.fid

Since 𝐹=𝐹, we have 𝐽=𝐽Δ𝑝=Δ𝑝Δ𝑝+Δ𝑝=0.

The last equation shows that there is no change in the total momentum of the two particles during the collision; the total momentum is constant. We say that momentum is a conserved quantity.

Law: Conservation of Momentum

For two or more bodies in an isolated system (which means that there are no external forces acting on it) acting upon each other, their total momentum remains constant: Δ𝑝=0.total

We now know that for any collision between two particles, the total momentum is conserved if we assume that there are no other interactions than the interaction between the colliding particles (we therefore assume there is no friction). However, does this mean that the total kinetic energy is conserved?

For the sake of simplicity, we will now consider collisions between particles that move on the same straight line. In this case, all vector quantities, such as velocity, force, and impulse, are one-dimensional vectors along the motion axis. This will allow us to use their single components along the motion axis instead of their vector forms in all equations.

Consider two bodies of equal mass, 𝑚, that move at equal speeds, 𝑣, toward each other along a smooth horizontal surface, approaching a position at which they will collide, as shown in the following figure.

The component of the initial (i.e., before collision) total momentum of the two bodies along the motion axis is given by 𝑝=𝑚𝑣+𝑚(𝑣)𝑝=𝑚𝑣𝑚𝑣=0.ii

Since the momentum of an isolated system is conserved, we know that, after collision, the total momentum of the two bodies is zero as well.

We see that it means that both velocities after collision must be of equal magnitude but opposite directions. The particles rebound on each other during collision (each particle is pushed back by the other particle); hence, their velocities after collision have changed sign and the magnitude of the velocities has possibly changed with respect to their precollision value, 𝑣. This can be represented as shown in the following diagram, where 𝑘 is a positive factor.

The total initial kinetic energy is 𝐾=12𝑚𝑣+12𝑚(𝑣)=𝑚𝑣.i

And the total final kinetic energy is 𝐾=12𝑚(𝑘𝑣)+12𝑚(𝑘𝑣)=𝑚𝑘𝑣𝐾=𝑘𝐾.ffi

Thus, 𝑘 is the ratio of the total kinetic energy of the bodies after and before the collision: 𝑘=𝐾𝐾.fi

As we are considering a closed system, the total mechanical energy cannot increase as energy cannot be created. The two particles move horizontally, which means that their potential gravitational energy is constant. Therefore, the total kinetic energy after the collision cannot exceed the total kinetic energy before the collision. Hence, we have 𝐾𝐾0𝑘10𝑘1.fi

When 𝑘=1, the magnitude of the velocities is conserved. Hence, we have 𝐾=𝐾.if

The kinetic energy is conserved in the collision; so, we say that the collision is elastic.

When 𝑘1, there is a loss of kinetic energy during the collision due to friction between the two particles. We say that a part of the kinetic energy has been dissipated. This type of collision is called an inelastic collision.

A perfectly inelastic collision is a collision during which the maximum amount of kinetic energy is lost. This happens when part of the kinetic energy is lost by bonding the bodies together. The two bodies therefore stick together after the collision and have the same final velocity.

In our example above, this would mean that both bodies have a zero velocity after the collision, since they need to have velocities of equal magnitude but opposite directions. We have then that 𝑘𝑣=𝑘𝑣=0 and 𝑘=0.

Let us recap what the different types of collision are.

Definition: Types of Collision of Particles

An elastic collision is a collision during which the kinetic energy is conserved.

An inelastic collision is a collision during which part of the kinetic energy is dissipated in friction between the two particles.

A perfectly inelastic collision occurs when kinetic energy is dissipated in friction and in bonding the two particles together, resulting in the two bodies sticking together and having the same velocity after collision.

Let us now consider a situation, again involving two equal-mass bodies, where the net momentum is nonzero, as shown in the following figure.

The component of the initial (i.e., before collision) total momentum of the two bodies along the motion axis is given by 𝑝=𝑚𝑣+0=𝑚𝑣.i

The following figure shows a valid postcollision outcome for the motion of the bodies.

For this outcome, the initially moving body is at rest while the initially at-rest body has a velocity equal to the precollision velocity of the initially moving body. The collision between the bodies is perfectly elastic as the kinetic energy is conserved 𝐾=𝐾=12𝑚𝑣if.

In the case of the perfectly inelastic collision, the two bodies stick together and have the same final velocity, as shown in the following figure. The component of the final velocity is 𝑣2 so that the total momentum is conserved.

While the initial net kinetic energy is 𝐾=12𝑚𝑣,i the final net kinetic energy is given by 𝐾=212𝑚𝑣2𝐾=14𝑚𝑣.ff

The loss of kinetic energy is Δ𝐾=|𝐾𝐾|Δ𝐾=|||14𝑚𝑣12𝑚𝑣|||Δ𝐾=14𝑚𝑣.fi

It can be mathematically shown that it is the maximum loss of kinetic energy while conserving the momentum.

The following figure shows a postcollision outcome that conserves momentum.

The initial momentum is still 𝑝=𝑚𝑣,i and the final momentum is 𝑝=𝑚(2𝑣)+𝑚(𝑣)=𝑚𝑣.f

The initial kinetic energy is given by 𝐾=12𝑚𝑣,i and the final kinetic energy is given by 𝐾=12𝑚(𝑣)+12𝑚(2𝑣)𝐾=12𝑚𝑣+2𝑚𝑣𝐾=52𝑚𝑣.fff

We see that 𝐾=5𝐾.fi

This result is not physically possible, however, as it would mean that energy is created. We see that conservation of momentum does not determine on its own whether a collision outcome is possible or not. It must be combined with the postcollision kinetic energy not exceeding its precollision value.

Let us look at an example of a collision between two bodies.

Example 1: Finding the Impulse of the Collision of Two Spheres in a Horizontal Plane

A sphere of mass 675 g was moving in a straight line on a smooth horizontal table at 31 cm/s. The sphere crashed into another smooth sphere of mass 837 g that was at rest on the table. If the first sphere came to rest as a result of impact, find the magnitude of the impulse between the two spheres.

Answer

The momentum of the initially moving sphere is given by 𝑝=𝑚𝑣,i where 𝑚 is the mass of the sphere and 𝑣 its velocity before collision: 𝑝=675(31)=20925/.igcms

The unit dyne is defined as dynegramcentimetresecond=.

So, the total momentum is 20‎ ‎925 dyn⋅s.

After the collision, the initially moving sphere is at rest. Its momentum after collision is therefore zero. The change in momentum for the sphere caused by the collision with another sphere is Δ𝑝=𝑝𝑝Δ𝑝=020925Δ𝑝=20925.fidyns

The impulse on this sphere caused by the force exerted by the other sphere during collision is given by 𝐽=Δ𝑝𝐽=20925.dyns

An impulse of equal magnitude but opposite direction is produced on the other sphere, initially at rest. Therefore, the magnitude of the impulse between the two spheres is |𝐽|=20925.dyns

Now let us see an example of how the conservation of momentum can determine the postcollision velocity of a body.

Example 2: Finding the Velocity of a Sphere after Collision with an Identical Sphere on the Same Line

Two spheres, 𝐴 and 𝐵, of equal mass were projected toward each other along a horizontal straight line at 19 cm/s and 29 cm/s respectively. As a result of the impact, sphere 𝐵 rebounded at 10 cm/s. Find the velocity of sphere 𝐴 after the collision given that its initial direction is the positive direction.

Answer

We can define the initial direction of sphere 𝐴 as positive. This makes the initial momentum of sphere 𝐴𝑝=19𝑚, and the initial momentum of sphere 𝐵𝑝=29𝑚, where 𝑚 is the mass of either sphere.

The net momentum is given by 𝑝=𝑝+𝑝=𝑚(1929)=𝑚(10)=10𝑚.net

After the collision, sphere 𝐵 has a reversed direction. Sphere 𝐵 was initially moving in the negative direction, so reversing makes it move in the positive direction. The momentum of sphere 𝐵 is given by 𝑝=10𝑚.

According to the principle of conservation of momentum, the momentum of sphere 𝐴 after the collision is given by 𝑝=𝑝𝑝,𝑝=10𝑚10𝑚=20𝑚.net

The velocity of sphere 𝐴 after the collision can be determined by the formula 𝑝=𝑚𝑣.

Making 𝑣 the subject gives 𝑣=20𝑚𝑚=20/.cms

Now let us see an example of how the conservation of momentum can determine the impulse of the force of the collision of bodies moving in opposite directions as well as the velocity of a body after the collision.

Example 3: Finding the Impulse Exerted on a Sphere Colliding with Another Sphere Moving in the Opposite Direction

Two spheres of masses 200 g and 350 g were moving toward each other along the same horizontal straight line. The first was moving at 14 m/s and the second at 3 m/s. The two spheres collided. As a result, the first sphere rebounded at 7 m/s in the opposite direction. Given that the positive direction is the direction of motion of the first sphere before the impact, determine the impulse 𝐼 the second sphere exerted on the first one and the speed 𝑣 of the second sphere after impact.

Answer

We will begin by converting the masses of the spheres to the SI base unit of mass, the kilogram.

The magnitudes of the impulses on the spheres are equal. As the velocity of the first sphere before and after the collision is known, the change in the velocity of the first sphere due to the collision can be determined: Δ𝑣=𝑣𝑣=7(14)=21/.ms

The impulse on the first sphere is equal to its change of momentum, so 𝐽=𝑚Δ𝑣=0.2(21)=4.2.Ns

An impulse of equal magnitude with the opposite sign acts on the second sphere. The impulse on the second sphere is added to its momentum. The initial momentum of the second sphere is given by the product of its mass and initial velocity. The initial velocity is in the negative direction: 𝑝=3𝑚,𝑝=0.35(3)=1.05/,iikgms and so the final momentum of the second sphere is 𝑝=1.05+4.2=3.15/.fkgms

The velocity of the second sphere after the collision is its momentum divided by its mass, which is 𝑣=𝑝0.35𝑣=3.150.35=9/.fffms

The speed of the second sphere after the collision is 9 m/s.

Now let us look at an example where two bodies collide and then move as a single body.

Example 4: Studying the Collision of Two Moving Bodies in the Same Line in Two Different Cases

Two spheres are moving along a straight line. One has a mass of 𝑚 and is moving at speed 𝑣, whereas the other one has a mass of 10 g and is moving at 36 cm/s. If the two spheres were moving in the same direction when they collided, they would coalesce into one body and move at 30 cm/s in the same direction. However, if they were moving in opposite directions, they would coalesce into one body which would move at 6 cm/s in the direction the first sphere had been traveling. Find 𝑚 and 𝑣.

Answer

The pre- and postcollision states of the two bodies when they have the same initial direction of motion are shown in the following figure.

Applying the law of conservation of total momentum of the system gives us 𝑚𝑣+36(10)=30(𝑚+10)𝑚𝑣+360=30𝑚+300𝑚𝑣+60=30𝑚60=30𝑚𝑚𝑣.

The pre- and postcollision states of the two bodies when they have opposite initial directions of motion are shown in the following figure.

Applying the law of conservation of total momentum of the system to this second situation gives us 36(10)+𝑚(𝑣)=6(𝑚+10)360𝑚𝑣=6𝑚60420=𝑚𝑣6𝑚.

We get the following system of equations: 60=30𝑚𝑚𝑣,420=𝑚𝑣6𝑚.

Adding both equations gives 60+420=30𝑚𝑚𝑣+𝑚𝑣6𝑚480=24𝑚𝑚=48024=20.g

The value of 𝑚 can be substituted into either equation of our system to obtain the value of 𝑣.

Using 60=30𝑚𝑚𝑣, we obtain 60=60020𝑣540=20𝑣𝑣=54020=27/.cms

Key Points

  • The momentum of a body is given by 𝑝=𝑚𝑣, where 𝑚 is the mass of the body and 𝑣 is the velocity of the body. For one-dimensional motion, it is sufficient to use the scalar form of momentum, 𝑝=𝑚𝑣, where one direction is taken as positive and the opposite direction is taken as negative.
  • The total momentum of particles in a closed system is a conserved quantity. Any change in the momentum of a body must be accompanied by a corresponding change in the momenta of some other bodies.
  • When two bodies collide, they exert forces of equal magnitude on each other for equal time intervals.
  • For a constant-mass body, the impulse produced by the action of a force over an interval of time equals the change in momentum of the body: 𝐽=𝐹𝑡=Δ𝑝.d
  • When two bodies collide, the magnitudes of the impulses on each body are equal.
  • A collision is perfectly elastic if the sum of the kinetic energies of the colliding bodies is the same before and after the collision.
  • The sum of the kinetic energies of the colliding bodies in a collision can decrease but it cannot increase.

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