# Impulse and Momentum | Physics for Engineers Lesson 3

Impulse and momentum are two important concepts in physics that are related to the motion of objects.

Momentum is the product of an object’s mass and velocity. It is a vector quantity, meaning that it has both magnitude and direction. The formula for momentum is:

p = m*v

where p is momentum, m is mass, and v is velocity.

Impulse is the change in momentum of an object over a period of time. It is also a vector quantity. The formula for impulse is:

J = Δp = mΔv

where J is impulse, Δp is the change in momentum, m is mass, and Δv is the change in velocity.

In other words, impulse is the force applied to an object over a certain amount of time that causes a change in its momentum. The greater the force and the longer the time it is applied, the greater the impulse.

The relationship between impulse and momentum is given by the impulse-momentum theorem, which states that:

J = Δp = FΔt

where F is the force applied, and Δt is the time interval over which the force is applied. This theorem shows that a change in momentum is directly proportional to the impulse applied to an object.

The conservation of momentum is another important concept related to momentum. In a closed system, the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction. This principle is known as the law of conservation of momentum.

In summary, momentum and impulse are important concepts in physics that describe the motion of objects and the forces that cause changes in their motion.

## Impulse

Impulse is a concept in physics that describes the change in momentum of an object over a period of time. It is defined as the product of the force applied to an object and the time interval over which it is applied. Impulse is a vector quantity, meaning that it has both magnitude and direction.

The formula for impulse is:

J = FΔt

where J is impulse, F is the force applied to the object, and Δt is the time interval over which the force is applied. The unit of impulse is newton-second (N-s).

The relationship between impulse and momentum is given by the impulse-momentum theorem, which states that the change in momentum of an object is equal to the impulse applied to it. Mathematically, this can be expressed as:

J = Δp

where J is impulse, and Δp is the change in momentum of the object.

The impulse-momentum theorem can be used to analyze the motion of objects in different situations. For example, in a collision between two objects, the impulse applied to each object can be calculated, and the resulting change in momentum can be determined. This can be used to predict the final velocities and directions of the objects after the collision.

The principle of conservation of momentum also applies to the concept of impulse. In a closed system, the total impulse applied to the system is equal to the total change in momentum of the system. This principle is based on the fact that the net external force acting on a closed system is zero, and therefore, the total momentum of the system is conserved.

In summary, impulse is a concept in physics that describes the change in momentum of an object as a result of a force acting on it over a period of time. It is a useful tool for analyzing the motion of objects in various situations, including collisions and interactions.

## Momentum

Momentum is a fundamental concept in physics that describes the motion of an object. It is defined as the product of an object’s mass and velocity. Momentum is a vector quantity, meaning that it has both magnitude and direction.

The formula for momentum is:

p = m * v

where p is momentum, m is mass, and v is velocity. The unit of momentum is kilogram-meter per second (kg m/s).

Momentum is conserved in a closed system, meaning that the total momentum of the system is constant. This principle is known as the law of conservation of momentum. It states that the total momentum of a closed system before an interaction or collision is equal to the total momentum after the interaction or collision.

The law of conservation of momentum can be used to analyze the motion of objects in various situations, including collisions and interactions. For example, in a collision between two objects, the total momentum of the system before the collision is equal to the total momentum of the system after the collision. This principle can be used to predict the final velocities and directions of the objects after the collision.

The concept of momentum is also related to the concept of impulse. Impulse is defined as the change in momentum of an object over a period of time. The impulse-momentum theorem states that the change in momentum of an object is equal to the impulse applied to it. This principle can be used to calculate the force applied to an object or the time required to change its momentum.

In summary, momentum is a fundamental concept in physics that describes the motion of an object. It is defined as the product of an object’s mass and velocity and is conserved in a closed system. The law of conservation of momentum and the impulse-momentum theorem are important tools for analyzing the motion of objects in various situations.

## Example

A 2 kg block is moving with a velocity of 10 m/s to the right. A force of 20 N is applied to the block for a time of 2 seconds in the opposite direction. What is the impulse applied to the block, and what is its final velocity?

Solution:

First, we can calculate the impulse applied to the block using the formula J = FΔt:

J = F * Δt = 20 N * 2 s = 40 N-s

The impulse applied to the block is 40 N-s.

Next, we can use the impulse-momentum theorem to calculate the change in momentum of the block:

J = Δp

where J is the impulse, and Δp is the change in momentum.

Δp = J = 40 N-s

Since the force is applied in the opposite direction to the initial velocity, the change in momentum is negative:

Δp = -20 kg m/s

Finally, we can use the equation for momentum to calculate the final velocity of the block:

p = m * v

where p is momentum, m is mass, and v is velocity.

The initial momentum of the block is:

p = m * v = 2 kg * 10 m/s = 20 kg m/s

The final momentum of the block is:

p’ = p + Δp = 20 kg m/s – 20 kg m/s = 0

Since the final momentum is zero, the final velocity of the block is also zero. Therefore, the block comes to a stop after the force is applied for 2 seconds.

So, the impulse applied to the block is 40 N-s, and its final velocity is zero.

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A force of 12 N is applied to a 3 kg object moving along the x-axis as a function of time given by F(t) = 3t^2 – 2t, where t is in seconds. The initial velocity of the object is 4 m/s to the right. Find the impulse applied to the object between t=0 and t=2 seconds.

Solution:

The impulse applied to the object can be found by integrating the force over the time interval from t=0 to t=2 seconds:

J = ∫ F(t) dt from t=0 to t=2

= ∫ (3t^2 – 2t) dt from t=0 to t=2

= [(t^3 – t^2) – (t^2 – t)] from t=0 to t=2

= (8 – 2/3) – (0 – 0) = 25/3 N-s

Therefore, the impulse applied to the object between t=0 and t=2 seconds is 25/3 N-s.

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A 1 kg ball is moving to the right with a velocity of 5 m/s. It collides with a stationary 2 kg ball. After the collision, the 1 kg ball moves to the left with a velocity of 3 m/s. Calculate the velocity of the 2 kg ball after the collision.

Solution:

We can use the law of conservation of momentum to solve this problem. According to this law, the total momentum of a system is conserved, meaning that the sum of the momenta of all the objects in the system remains constant before and after a collision.

Before the collision, the 1 kg ball has momentum to the right, while the 2 kg ball is at rest. Therefore, the total momentum of the system is:

p_total = m1v1 + m2v2 = 1 kg * 5 m/s + 2 kg * 0 m/s = 5 kg m/s to the right

After the collision, the 1 kg ball moves to the left, and the 2 kg ball moves to the right. Let’s assume that the velocity of the 2 kg ball after the collision is v2′. The total momentum of the system after the collision is:

p_total’ = m1v1′ + m2v2′ = 1 kg * (-3 m/s) + 2 kg * v2′

By the law of conservation of momentum, the total momentum before and after the collision must be equal:

p_total = p_total’

Substituting the values we found earlier, we get:

5 kg m/s to the right = 1 kg * (-3 m/s) + 2 kg * v2′

Solving for v2′, we get:

v2′ = (5 kg m/s + 3 kg m/s) / 2 kg = 4 m/s to the right

Therefore, the velocity of the 2 kg ball after the collision is 4 m/s to the right.

In summary, we can use the law of conservation of momentum to calculate the velocity of an object after a collision, based on the momentum of the system before the collision and the masses and velocities of the objects involved. In this case, the momentum of the system before the collision was conserved, allowing us to solve for the velocity of the 2 kg ball after the collision.

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A 1 kg ball is moving to the right with a velocity of 5 m/s. It collides with a stationary 2 kg ball. After the collision, the 1 kg ball moves to the left with a velocity of 3 m/s. Calculate the velocity of the 2 kg ball after the collision, using calculus.

Solution:

We can use the law of conservation of momentum to solve this problem, which states that the total momentum of a system is conserved. Therefore, the sum of the momenta of all the objects in the system before and after the collision is the same. Mathematically, we can express this as:

m1v1 + m2v2 = m1v1′ + m2v2′

where m1 and m2 are the masses of the two balls, v1 and v2 are their velocities before the collision, and v1′ and v2′ are their velocities after the collision.

We can solve for the velocity of the 2 kg ball after the collision, v2′, using calculus. First, let’s find the momentum of each ball before and after the collision:

Before the collision:

p1 = m1v1 = 1 kg * 5 m/s = 5 kg m/s to the right p2 = m2v2 = 2 kg * 0 m/s = 0 kg m/s

Total momentum: p_total = p1 + p2 = 5 kg m/s to the right

After the collision:

p1′ = m1v1′ = 1 kg * (-3 m/s) = -3 kg m/s to the left p2′ = m2v2′

Now we can solve for v2′ using the law of conservation of momentum:

m1v1 + m2v2 = m1v1′ + m2v2′

1 kg * 5 m/s + 2 kg * 0 m/s = 1 kg * (-3 m/s) + 2 kg * v2′

5 kg m/s = -3 kg m/s + 2 kg * v2′

8 kg m/s = 2 kg * v2′

v2′ = 4 m/s to the right

Therefore, the velocity of the 2 kg ball after the collision is 4 m/s to the right.

In summary, we used calculus to solve for the velocity of the 2 kg ball after the collision, by applying the law of conservation of momentum and solving for the unknown variable using algebraic manipulation.

## Exercises

1. A 0.5 kg object is moving to the right with a velocity of 10 m/s. It collides with a stationary 1 kg object. After the collision, the 0.5 kg object moves to the left with a velocity of 2 m/s. What is the velocity of the 1 kg object after the collision? Answer: 6 m/s to the right.
2. A 2 kg ball is rolling to the right with a velocity of 3 m/s. It collides with a stationary 1 kg ball. After the collision, the 2 kg ball moves to the left with a velocity of 1 m/s, and the 1 kg ball moves to the right with a velocity of 4 m/s. Calculate the velocity of the center of mass of the system after the collision. Answer: 1 m/s to the right.
3. A 0.1 kg bullet is fired from a gun with a velocity of 500 m/s. The bullet embeds itself in a 5 kg block of wood initially at rest. What is the velocity of the block of wood after the bullet embeds itself in it? Answer: 10 m/s to the left.

## Quiz

1. A 5 kg ball is moving to the left with a velocity of 4 m/s. It collides with a stationary 2 kg ball. After the collision, the 5 kg ball moves to the right with a velocity of 2 m/s. What is the velocity of the 2 kg ball after the collision?
2. A 0.5 kg object is moving to the right with a velocity of 20 m/s. It collides with a stationary 1 kg object. After the collision, the 0.5 kg object moves to the left with a velocity of 5 m/s. What is the velocity of the 1 kg object after the collision?
3. A 2 kg ball is moving to the right with a velocity of 6 m/s. It collides with a stationary 1 kg ball. After the collision, the 2 kg ball moves to the left with a velocity of 3 m/s, and the 1 kg ball moves to the right with a velocity of 8 m/s. Calculate the velocity of the center of mass of the system after the collision.