Intro to Impulse: Videos & Practice Problems

Impulse (J) is defined as the change in momentum (Δp), expressed as J = Δp. This can be rewritten using Newton's second law as J = FΔt, where F is force and Δt is the time interval. The relationship between impulse and momentum is crucial, as it shows how forces acting over time can change an object's momentum. For example, applying a 100 N force for 8 seconds results in an impulse of 800 N·s, leading to a final velocity of 16 m/s for a crate initially at rest. Understanding this relationship is essential for solving physics problems involving motion.

Impulse & Impulse-Momentum Theorem

Impulse is a crucial physical quantity that relates directly to momentum, defined as the change in momentum experienced by an object when it undergoes an impulse. This relationship can be expressed mathematically as:

$$ J = \Delta P $$

Here, \( J \) represents impulse, and \( \Delta P \) denotes the change in momentum. To further understand this, we can break down momentum, which is given by the formula:

$$ P = m v $$

where \( m \) is mass and \( v \) is velocity. The change in momentum, \( \Delta P \), can be expressed as:

$$ \Delta P = m v_{\text{final}} - m v_{\text{initial}} $$

To connect impulse with Newton's second law, we start with the equation:

$$ F = m a $$

Acceleration can be rewritten in terms of change in velocity over change in time:

$$ a = \frac{\Delta v}{\Delta t} $$

Substituting this into the force equation gives:

$$ F = \frac{\Delta P}{\Delta t} $$

From this, we can derive the impulse equation by rearranging it to:

$$ J = F \Delta t $$

Impulse is thus defined as the product of force and the time duration over which the force acts. The units of impulse are Newton-seconds (N·s), which are equivalent to the units of momentum, kilogram meters per second (kg·m/s).

To illustrate the application of impulse, consider a scenario where a crate is initially at rest and is pushed with a force of 100 Newtons for 8 seconds. The impulse delivered to the crate can be calculated as:

$$ J = F \Delta t = 100 \, \text{N} \times 8 \, \text{s} = 800 \, \text{N·s} $$

Next, to find the final speed of the crate after the force has been applied, we can use the impulse-momentum theorem:

$$ J = m v_{\text{final}} - m v_{\text{initial}} $$

Given that the initial velocity is 0, we can simplify this to:

$$ 800 = m v_{\text{final}} $$

If the mass of the crate is 50 kg, we can solve for the final velocity:

$$ v_{\text{final}} = \frac{800}{50} = 16 \, \text{m/s} $$

This demonstrates how impulse can be used to determine changes in velocity when a constant force is applied over a specific time interval.

Moreover, it is important to note the similarities between impulse and momentum, as well as between work and kinetic energy. Just as impulse (force times time) results in a change in momentum, work (force times displacement) results in a change in kinetic energy. This parallel highlights the interconnectedness of these fundamental concepts in physics.

You throw a 100-g ball with 30m/s. If the ball is in your hand for 0.2s during the throw, a) calculate the impulse
you deliver to it. b) Calculate the average Force that you exert on the ball.

• A. $J=3,000\operatorname{kg}$•m/s, $F_{avg}=15,000N$
• B. $J=3\operatorname{kg}$•m/s, $F_{avg}=15N$
• C. $J=0.2\operatorname{kg}$•m/s, $F_{avg}=1N$
• D. $J=2\operatorname{kg}$•m/s, $F_{avg}=10N$

Answer: (B)

Impulse between Bouncy Ball & Wall

In this scenario, we analyze the behavior of a bouncy ball with a mass of 0.15 kg that strikes a wall while traveling at an initial speed of 40 meters per second to the right. Upon collision, the ball rebounds in the opposite direction, resulting in a final velocity of -45 meters per second, indicating a leftward motion. It is crucial to maintain a consistent direction for positive and negative velocities throughout the calculations.

To determine the impulse (J) delivered to the ball during the collision, we can utilize the relationship between impulse and momentum. The impulse can be calculated using the formula:

J = m(v_{final} - v_{initial})

Substituting the known values, we have:

J = 0.15 \, \text{kg} \times (-45 \, \text{m/s} - 40 \, \text{m/s})

Calculating this gives:

J = 0.15 \, \text{kg} \times (-85 \, \text{m/s}) = -12.75 \, \text{kg m/s}

This negative value indicates that the impulse acts in the leftward direction, consistent with the change in momentum caused by the collision.

In the second part of the problem, we need to find the duration of contact (Δt) between the ball and the wall. We know the average force exerted by the wall during the bounce is 410 newtons, which we will consider as negative to align with our chosen direction. The impulse can also be expressed as:

J = F \cdot \Delta t

Substituting the known values, we have:

-12.75 \, \text{kg m/s} = -410 \, \text{N} \cdot \Delta t

To solve for Δt, we rearrange the equation:

\Delta t = \frac{-12.75}{-410} \approx 0.031 \, \text{s}

This result indicates that the ball is in contact with the wall for approximately 31 milliseconds during the collision. Understanding these concepts of impulse and momentum is essential in analyzing collisions and the forces involved in such interactions.

You catch a 0.6 kg ball initially moving with 10 m/s. Calculate the impulse delivered to the ball during the catch.

• A. $16N\cdot s$
• B. $6N\cdot s$
• C. $0.6N\cdot s$
• D. $60N\cdot s$

Answer: (B)

Car Colliding with Wall

In this scenario, we analyze a 1200-kilogram car colliding with a wall at an initial speed of 20 m/s. During the collision, the car crumples over a distance of 1 meter. To determine the duration of the collision, we utilize the impulse-momentum theorem, which states that the impulse (J) is equal to the change in momentum (Δp). The impulse can be expressed as:

$$ J = F \cdot \Delta t = m \cdot (v_{final} - v_{initial}) $$

Here, we need to find the time of collision (Δt). However, since both the force (F) and time (Δt) are unknown, we cannot directly solve for Δt using this equation. Instead, we assume constant acceleration during the collision, allowing us to apply kinematic equations.

We can use the equation:

$$ \Delta x = \frac{(v_{initial} + v_{final})}{2} \cdot \Delta t $$

Given that:

• Δx = 1 m (the distance crumpled)
• v_initial = 20 m/s
• v_final = 0 m/s

Substituting these values into the equation gives:

$$ 1 = \frac{(20 + 0)}{2} \cdot \Delta t $$

Solving for Δt results in:

$$ \Delta t = \frac{1}{10} = 0.1 \text{ seconds} $$

Now that we have the time of collision, we can find the average force exerted by the wall on the car. Using the impulse equation again, we can rearrange it to solve for force:

$$ F = \frac{J}{\Delta t} $$

Substituting the change in momentum:

$$ F = \frac{m \cdot (v_{final} - v_{initial})}{\Delta t} $$

Plugging in the known values:

$$ F = \frac{1200 \cdot (0 - 20)}{0.1} $$

This simplifies to:

$$ F = \frac{-24000}{0.1} = -240000 \text{ N} $$

Since we are interested in the magnitude of the force, we report it as:

$$ |F| = 240000 \text{ N} $$

This substantial force is necessary to stop a heavy car moving at 20 m/s over a very short time frame of 0.1 seconds, illustrating the significant impact forces involved in collisions.

Here’s what students ask on this topic:

What is the definition of impulse in physics?

Impulse in physics is defined as the change in momentum of an object when a force is applied over a specific time interval. Mathematically, it is expressed as:

$J=\mathrm{\Delta p}$

where $J$ is the impulse and $\mathrm{\Delta p}$ is the change in momentum. It can also be written as:

$J=F\mathrm{\Delta t}$

where $F$ is the force applied and $\mathrm{\Delta t}$ is the time interval over which the force is applied. The units of impulse are Newton-seconds (N·s) or kilogram meters per second (kg·m/s).

How is impulse related to momentum?

Impulse is directly related to momentum through the equation:

$J=\mathrm{\Delta p}$

This means that the impulse experienced by an object is equal to the change in its momentum. Momentum (p) is defined as the product of an object's mass (m) and its velocity (v), so a change in momentum can be expressed as:

$\mathrm{\Delta p}=m\mathrm{\Delta v}$

where $\mathrm{\Delta v}$ is the change in velocity. Therefore, when a force is applied over a time interval, it changes the object's velocity, thereby changing its momentum. This relationship is crucial for solving problems involving motion and forces in physics.

What are the units of impulse?

The units of impulse are Newton-seconds (N·s) or kilogram meters per second (kg·m/s). These units come from the definition of impulse as the product of force (F) and time (Δt):

$J=F\mathrm{\Delta t}$

Since force is measured in Newtons (N) and time in seconds (s), the resulting unit is Newton-seconds (N·s). Alternatively, since impulse is also the change in momentum, and momentum is measured in kilogram meters per second (kg·m/s), impulse can also be expressed in these units.

How do you calculate the final velocity of an object using impulse?

To calculate the final velocity of an object using impulse, you can use the impulse-momentum theorem:

$J=mv$_f-mv_i

where $J$ is the impulse, $m$ is the mass, $v$_f is the final velocity, and $v$_i is the initial velocity. Rearrange the equation to solve for the final velocity:

$v$_f=Jm+v_i

For example, if a 50 kg crate initially at rest (vi = 0) experiences an impulse of 800 N·s, the final velocity (vf) is:

$v$_f=80050=16 m/s

What is the impulse-momentum theorem?

The impulse-momentum theorem states that the impulse applied to an object is equal to the change in its momentum. Mathematically, it is expressed as:

$J=\mathrm{\Delta p}$

or

$F\mathrm{\Delta t}=mv$_f-mv_i

where $J$ is the impulse, $F$ is the force applied, $\mathrm{\Delta t}$ is the time interval, $m$ is the mass, $v$_f is the final velocity, and $v$_i is the initial velocity. This theorem is crucial for understanding how forces affect the motion of objects over time.