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

# Lesson Video: Collisions and Conservation of Momentum

In this video, 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.

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### Video Transcript

In this video, 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. We will begin by looking at some key definitions and formulae that we will need in this video.

We recall that the momentum of a body is equal to its mass multiplied by its velocity. This leads us to the formula π is equal to π multiplied by π£, where π is the momentum of a body. In this video, the mass will be measured in either kilograms or grams. The velocity of a body will be measured in meters per second or centimeters per second. The units of momentum will therefore be kilogram meters per second or gram centimeters per second. The impulse of a force is equal to the change in momentum of a body. It is also equal to the magnitude of the force multiplied by the length of time the force is applied.

This gives us the following formula. The impulse is equal to the force multiplied by the time which is equal to the mass multiplied by the final velocity minus the mass multiplied by the initial velocity. These quantities are scalars and not vectors. This change in momentum is often referred to as the momentum after the collision minus the momentum before the collision. The units of impulse are newton seconds or dyne seconds. These are dimensionally equivalent to the momentum units. This means that newton seconds and kilogram meters per second are interchangeable. The same is true of dyne seconds and gram centimeters per second. One newton second corresponds to one newton force applied for one second.

We will now look at the conservation of momentum and how it can be applied to elastic and inelastic collisions in one dimension. The conservation of momentum states that the total momentum of two objects before the collision is equal to the total momentum of the objects after the collision. We can demonstrate this using before and after diagrams. Letβs assume we have two spheres of mass π one and π two traveling in the same direction with velocities π’ one and π’ two. The two spheres collide and after the collision are traveling with velocities π£ one and π£ two, respectively. We know the momentum before must be equal to the momentum after. As momentum is equal to mass multiplied by velocity, π one π’ one plus π two π’ two must be equal to π one π£ one plus π two π£ two. We will use this formula to calculate unknown velocities or masses in problems involving collisions.

Before moving on to some questions, weβll briefly consider the difference between elastic and inelastic collisions. An elastic collision is one where there is no net loss of kinetic energy as a result of the collision. In reality, examples of perfectly elastic collisions are not part of our everyday experience. However, there are some examples of collisions in mechanics where the energy lost can be negligible. These collisions can be considered elastic, even though they are not perfectly elastic. Examples include collisions of pool balls or the balls in a Newtonβs cradle. When two pool balls collide, the conservation of momentum holds. The momentum before the collision is equal to the momentum after the collision.

An inelastic collision is one where there is a loss of kinetic energy. When two bodies coalesce, which means they join to make one body, the collision is perfectly inelastic. Whilst in the real world, most collisions are somewhere in between perfectly elastic and perfectly inelastic, in this video, we will deal with the two extremes. In our first question, we will find the velocity of a sphere after a collision.

Two spheres π΄ and π΅ of equal mass were projected toward each other along a horizontal straight line at 19 centimeters per second and 29 centimeters per second, respectively. As a result of the impact, sphere π΅ rebounded at 10 centimeters per second. Find the velocity of sphere π΄ after the collision given that its initial direction is the positive direction.

In order to answer this question, we will draw two diagrams modeling the situation before and after the collision. Before the collision, the two spheres of equal mass π are moving toward each other, sphere π΄ at 19 centimeters per second and sphere π΅ at 29 centimeters per second. We are told that after the collision, sphere π΅ rebounds at 10 centimeters per second. And we need to calculate the velocity of sphere π΄. Weβre also told that sphere π΄ was initially moving in the positive direction. In order to answer this question, we will use the conservation of momentum. This states that the momentum before is equal to the momentum after.

As momentum is equal to mass multiplied by velocity, our equation is π one π’ one plus π two π’ two is equal to π one π£ one plus π two π£ two. Substituting in the initial velocities, we have π multiplied by 19 plus π multiplied by negative 29. This is because sphere π΅ is moving in the negative direction. This is equal to π multiplied by π£, the velocity of sphere π΄ after the collision, plus π multiplied by 10.

As the mass of both spheres is the same and cannot be equal to zero, we can divide through by the variable π. 19 plus negative 29 is equal to negative 10. This leaves us with negative 10 is equal to π£ plus 10. We can then subtract 10 from both sides of this equation giving us π£ is equal to negative 20. The velocity of sphere π΄ after the collision is negative 20 centimeters per second. This means that it is moving with a speed of 20 centimeters per second in the negative direction.

In our next question, we need to find the impulse exerted on a sphere.

Two spheres of masses 200 grams and 350 grams were moving toward each other along the same horizontal straight line. The first was moving at 14 meters per second and the second at three meters per second. The two spheres collided. As a result, the first sphere rebounded at seven meters per second in the opposite direction. Given that the positive direction is the direction of motion of the first sphere before the impact, determine the impulse πΌ the second sphere exerted on the first one and the speed π£ of the second sphere after impact.

In order to answer this question, we will draw two diagrams showing what is happening before and after the impact. We do need to be careful here when looking at the units. The masses are given in grams and the velocities in meters per second. When dealing in meters per second, we need the masses to be in kilograms. Likewise, if the velocities were in centimeters per second, we would need the masses to be in grams. We know that 1000 grams is equal to one kilogram. This means that the masses of the two spheres are 0.2 kilograms and 0.35 kilograms, respectively. It is important to note that the first sphere is moving in the positive direction. This means that the second sphere is moving in the negative direction, as it is moving towards it.

After the collision, we are told that the first sphere rebounds at seven meters per second in the opposite direction. The speed of the second sphere after impact is π£. The first part of our question wants us to calculate the impulse that the second sphere exerted on the first one. We know that impulse is equal to the change in momentum. This leads us to the equation πΌ is equal to ππ£ minus ππ’ as momentum is equal to the mass multiplied by the velocity. This can also be written as π multiplied by π£ minus π’ by factoring out the mass.

We know that the mass of sphere π΄ is 0.2 kilograms. The velocity after the collision is negative seven as it is moving at a speed of seven meters per second in the negative direction. The initial velocity before the collision was 14 meters per second. The impulse is therefore equal to 0.2 multiplied by negative 21. This gives us an answer of negative 4.2. The impulse that the second sphere exerts on the first one is negative 4.2 newton seconds. We recall that our units for impulse are newton seconds when the mass is in kilograms and the velocity is in meters per second.

The second part of this question wants us to calculate the speed π£. Using the conservation of momentum, we know that the momentum before is equal to the momentum after. This leads us to the equation π one π’ one plus π two π’ two is equal to π one π£ one plus π two π£ two. Substituting our values for the masses and velocities gives us 0.2 multiplied by 14 plus 0.35 multiplied by negative three is equal to 0.2 multiplied by negative seven plus 0.35 multiplied by π£. The left-hand side simplifies to 1.75. And on the right, we have negative 1.4 plus 0.35π£. Adding 1.4 to both sides gives us 3.15 is equal to 0.35π£. Finally, dividing both sides by 0.35 gives us π£ is equal to nine. The speed of the second sphere after impact is nine meters per second. It is also worth noting that this sphere is traveling in the positive direction.

In the final question of this video, we will deal with an inelastic collision.

Two spheres are moving along a straight line. One has mass π and is moving at speed π£, whereas the other has a mass of 10 grams and is moving at 36 centimeters per second. If the two spheres were moving in the same direction when they collided, they would coalesce into one body and move at 30 centimeters per second in the same direction. However, if they were moving in opposite directions, they would coalesce into one body, which would move at six centimeters per second in the direction the first sphere had been traveling. Find π and π£.

There are two scenarios in this question, and in both cases, the bodies coalesce. This means that they join together and the collision is inelastic. In the first scenario, the two bodies are moving in the same direction. In this situation, they join together and move at a speed of 30 centimeters per second. In the second scenario, they were originally moving towards each other. And they end up moving in the same direction as the first sphere with a speed of six centimeters per second. In order to answer this question, we will use the conservation of momentum. This states that the momentum before is equal to the momentum after. The formula we will use is π one π’ one plus π two π’ two is equal to π one π£ one plus π two π£ two.

We will now clear some space to answer the question. As the spheres are joining together, we will only have one product on the right-hand side. In our first scenario, we have ππ£ plus 10 multiplied by 36 is equal to π plus 10 multiplied by 30. Distributing the parentheses gives us 30π plus 300. We can then subtract 360 from both sides so that ππ£ is equal to 30π minus 60. We repeat this for the second scenario. The only differences are the 36 is now negative as the second sphere is moving in the opposite direction and the final velocity is six. This simplifies to ππ£ minus 360 is equal to six π plus 60. This time, we can add 360 to both sides, such that ππ£ is equal to six π plus 420.

We now have a pair of simultaneous equations where the left-hand side of both of them is ππ£. This means that 30π minus 60 must be equal to six π plus 420. Adding 60 and subtracting six π from both sides gives us 24π is equal to 480. We can then divide both sides by 24 giving us π is equal to 20. The mass of the first sphere is 20 grams. We can now substitute this into one of our equations. We will choose equation one. This gives us 20π£ is equal to 30 multiplied by 20 minus 60. The right-hand side simplifies to 540. We can then divide both sides of the equation by 20 giving us π£ is equal to 27. The initial velocity of the first sphere is 27 centimeters per second. We have now found the mass and speed of the first sphere that would result in the two spheres coalescing on collision.

We will now summarize the key points from this video. In this video, we saw that momentum is equal to mass multiplied by velocity. We also saw that from the conservation of momentum, the momentum before a collision is equal to the momentum after. This led us to the equation, π one π’ one plus π two π’ two is equal to π one π£ one plus π two π£ two, where π one and π two are the masses of the two bodies. π’ one and π’ two are the velocities before the collision, and π£ one and π£ two are the velocities after the collision.

We also saw that impulse is equal to the change in momentum. This gave us the equation πΌ is equal to ππ£ minus ππ’. It is the momentum after the collision minus the momentum before. For the questions in this video, we model the situation as perfectly elastic or perfectly inelastic. This means that no energy was lost in the collision. In a perfectly inelastic collision, we know that the two bodies will coalesce or join together.

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