Unit 4 (4.1 & 4.2)

Unit 4.1

Linear Momentum

Linear Momentum: \overrightarrow{p}=m\overrightarrow{v}

Unit: kg * m/s

Inertia: The tendency to resist motion, the object with the greater mass has the most inertia

Momentum is a vector quantity

Total Momentum is calculated using the SUMs

Example:

Solution

All objects have the same momentum (6 kg*m/s)

The 3.0 kg ball has the most inertia

Force and Momentum Change

Newton’s Second Law: When a net force acts on an object, the velocity of the object changes

(an object in motion stays in motion unless affected by a net force)

\sum_{}^{} \overrightarrow{F}=m\cdot\overrightarrow{a}

Relating Force to Momentum

Final Equation:\overrightarrow{F_{avg}}=\frac{dp}{dt}

Example 1

Solution

B

Example 2

Solution

C

We are taking the magnitude of the system’s momentum

***Momentum is a vector!!

Taking the magnitude of a vector is used by the formula

\sqrt{x^2+y^2}=v^2

Example 3:

Solution

D

Relate Kinetic Energy with Momentum

First Way:

Final Equation: KE = \frac12PV

• The object with more velocity has MORE kinetic energy IF momentum is constant

Second Way:

Final Equation: KE = \frac{p^2}{2m}

• Mass is inversely proportional to kinetic energy IF momentum is constant

Example 3

Solution

A

F_net = \frac{dp}{dt} , so Force is the DERIVATIVE OF MOMENTUM

Unit 4.2

Impulse-Momentum Theorem

Impulse: J = \int_{}^{}fd\left(t\right) OR \Delta p OR f_{avg}\Delta t\!

Unit: N\cdot s OR kg * m/s

Example 1

Solution

C

To calculate the change of momentum we look at Impulse, since we have an Force v Time graph

Impulse = area under the curve of a force v time graph, AND change in momentum

• so for the interval (0,2), the area under the graph is 20 N*s

• for the interval (2,4), the area under the graph is -20 N*s

To find the TOTAL Momentum, we add the momentum together

• 20 + (-20) = 0

Example 2a

Solution

B

Force and Time is constant for both blocks, so using impulse-momentum theorem

J = f_{avg}\Delta t\! , so Impulse is constant

If impulse is constant then Momentum is ALSO constant

Example 2b

Solution

Now that we have Force and distance, we cannot use impulse-momentum theorem anymore

Instead we will use the Work-Energy Theorem

W = \int_{}^{}f\left(x\right)dx OR \Delta K

If we know that Force and Distance are both constant, then the Work & KE are constant as well

We can relate KE to momentum with the formula: KE = \frac{p^{\circ}}{2m}

• KE = \frac{p^{\circ}}{2m} ——> KE(2m) = p²

If KE is constant in both scenarios, then when comparing the two momentums, we can eliminate KE from the equation SO

• 2m = p²

  • If Scenario A has mass m, then p² = 2m

  • If Scenario B has mass 4m, then p² = 8m

Taking the square root of both we get

• Scenario A = \sqrt{2m}

• Scenario B = \sqrt{8m} OR this can be rewritten as 2\sqrt{2m}

So Scenario B has TWICE the change of momentum than scenario A does

Difference Between Theorems

Impulse-Momentum Theorem:

change in object’s momentum is due to impulse which depends on force AND TIME

• J=\Delta p

• J = \int_{}^{}fd\left(t\right)

Work-Energy Theorem

Change in object’s KE is due to work which depends on force and DISTANCE

• W = \int_{}^{}f\left(x\right)dx

• W = \Delta K

Example 3

Solution

B

Knowing that we are given the force v time function and time, we can use the Impulse-Momentum Theorem

Impulse = \int_{}^{}fd\left(t\right)

The integral of 0.5t ———> 0.25t²

Inputting values we find that J = 4,

• since Impulse = \Delta p , we know \Delta p = 4

\Delta p = mv, inputting values we find that 4 = 5v. so v = \frac45 OR 0.8

Example 4a

Solution

1 N*s or 1 kg*m/s

If we know that Impulse-Momentum Theorem is viable then using J = \Delta p and F_{avg}=\frac{\Delta p}{\Delta t} best

We know that J = 4, so \Delta p = 4

• Inputting values into F_{avg}=\frac{\Delta p}{\Delta t} ————> F_avg = 4/4, SO

F_{avg = 1 N*S}

Example 4b

Solution

p = 6.3 kg *m/s

We cannot use Impulse-Momentum theorem here, since there is no time variable

Since there is a Force v position function, we use the Work-Energy Theorem

• W=\int_{}^{}F\left(x\right)dx

• W = \Delta K

Integrating F_x = 0.5x ———→ F_x = \frac{x^2}{4}

Putting values we get F_x = 4, so W = 4

Relating KE to momentum we get: KE = \frac{p^2}{2m}

• 4 = \frac{p^2}{2m} ———> 40 = p²

• p = 6.3 kg *m/s

Collision Time and Impact Force

WITHOUT increasing Force

You can increase an object’s change in momentum by increasing the collision time

You can calculate the Collision Time:

1.) using a Force v Time graph by looking directly at the time-axis

2.) Using Momentum and the Average Force

• Impulse =\Deltap and Impulse =F_{avg}\Delta t

• So \Deltap =F_{avg}\Delta t , to isolate time we divide \Deltap byF_{avg} (\frac{\Delta p}{F_{avg}} = \Delta t )

Example 1

Solution

Using Energy Conservation Theorem, we can calculate the velocity and as such the momentum using what is given

Finding change in Momentum from drop height & rebound height:

• PE = KE ————————→ mgh = \frac12mv^2

• Isolate velocity: \sqrt{2gh} = v

• For Initial velocity, Since the ball is going down, height is negative

• m_1v_{f}-m_1v_{i} = \Delta p

using a Scale:

• Find out the Normal Force of the ball to find out the F_net

• The Free-Body diagram reveals F_g and F_N for the Y-Component

  • F_NET = f_N - f_g

  • So Divide the Force by the change in momentum, (\frac{\Delta p}{F_{avg}} ), we found by using velocity (m_1v_{f}-m_1v_{i} = \Delta p ), gives us collision Time

Example 2

Solution

Both balls experiences the same impulse

Both balls drop from the same height and rebound to half the drop height, so the change in Kinetic Energy, i.e. change in velocity is the same

Since p = mv, then the \Delta p is the same as mass is constant

Case 1 experiences the greater average impulse force

Dropping a ball on a rubber floor makes the collision time longer

• Generally rubber and stretchy materials decrease impact force and increase collision time

• As a result, if momentum is constant between the two cases, then the impulse force of Case 2 must be less, as Case 2’s collision time is longer

  • \Delta p =F_{avg}\Delta t

Example 3

Solution

Both of the cars have equal impulse since their start and end momentum is the same, their change in momentum is the same, as a result, their impulse is the same

• J=\Delta p=m\left(v_{f}-v_{i}\right)

The car hitting the brick wall experiences a greater average impact force

• The brick wall decrease the collision speed, since you stop almost as soon as you hit it

• so the average impact force needs to be higher

The opposite of increasing collision time decreases average impact force is why wearing your seat belt is important

Example 4

Solution

B

The impulse is equal in both cases as they were dropped from the same height and both ended up stopping in some way

However, since the carpet increases the collision time, its impact force is less on the glass

Example 5

Solution

Knees bent

Impulse is equal in both cases

It takes more time for the girl’s center off mass to come to a stop with her knees bent so less average impact force occurs

Bouncing

The greater the rebound effect, the greater the change in velocity & momentum

• Greater Average Impulse Force

• Greater acceleration, etc

Example 1

Solution

Case 1 has the greater impulse

Case 1 rebounds, the collision has a change in direction of motion

• velocity change & momentum change is greater

Case 1 has the greater impact force

• In Case 2, the crumbling nature of the collision extends the collision time

  • also minimizes the change in impulse and momentum

The slope of Case 1 is steeper which implies a greater slope

• the derivative of velocity is acceleration, so the greater slope implies greater acceleration

• Acceleration implies greater average impulse force

  • F_{net}=ma

Example 2

Solution

A

Bounding results in greater change in momentum and velocity

• creates greater impulse

Given they both have equal collision time, the bouncing ball has greater average impulse force

Example 3

Solution

A

The magnitude of momentum change is greater for case 1, so if they both have equal average impulse force, then Case 1 has greater collision time

• J=F_{avg}\Delta t=\Delta p

• m₁v_f - m₁v_{i = \Delta p}

  • Case 1: mv - (-mv) = 2mv

    • The ball is perfectly elastic: all KE is preserved, so the rebound and initial velocity are equal

    • V initial is negative relative to V final because theyre in opposite directions & velocity is a vector

  • Case 2: m\cos\theta v - (-m\cos\theta v) = 2m\cos\theta v

Example 4

MultiPart Examples

Ball and Block (Unit 1 + Unit 4)

Solution

a.) 12 N*s

b.) v_b = 2.4 m/s

c.) v_c = 2 m/s

d.) 0.2 m

a.) 12 kg * m/s

Since we are given a Force v time graph, we’ll use the formula Impulse = \int Fdt

b.) v_b = 2.4 m/s

To find the velocity after the collision of the ball, we use the fact that \Delta p=\int fdt

• According to part A, Impulse = 12, so \Delta p = 12

\Delta p=m\left(v_{f}-v_{i}\right)

Since the ball was initially at rest, v_i = 0, so

12 = 5(v_f) ————> v_f = 2.4

c.) v_c = 2 m/s

According to Newton’s second law F_AB = -F_BA, so the force done onto the ball over the collision time has the same magnitude as the force done on the block.

• Using that we can find the Impulse is now -12

Using \Delta p=m\left(v_{f}-v_{i}\right) ,

we get -12 = 0.5(v_f - 26) ———→ -24 = v_f - 26 ————→ v_f = 2

d.) 0.2 m

Knowing the distance from the ground gives us a possibility if using kinematics to find the horizontal distance each ball goes.

• D = v_i(t) - \frac12 (a)t²

  • Under this Tom & Jerry problem, d_x = v_i(t) & d_y = -\frac12 (a)t²

So we find time using the formula we have for d_x,

• 1.2 = -\frac12 (10)t²———→ \sqrt{0.24} OR 0.50 sec

• Plugging that in the d_x formulas for both ball and block

  • Block: 1 m

  • Ball 1.2 m

The distance between the two is 0.2 m

Toy Rocket Launch

Solution

a.) Impulse_net = 18 N*s

J = \int fdt

F_net = ma

we can separate Force into mass * acceleration

• J = m\cdot\int adt

• Since we have the equation for acceleration, we can integrate that and then multiply by mass

a = 9 - 0.25t² ——→ a. = 9t - \frac{0.25t^3}{3} OR 9t-\frac{t^3}{12}

b.) v_f at 6 seconds is 36 m/s

c.)

We already know one point from part a and b, so we can plot, (6, 36).

Rocket still goes up even though the fuel is exhausted due to momentum & inertia

• eventually reaches top height, max

velocity decreases from (6, T_top) because the acceleration is less, so the velocity is less

From (T_top, \infty ), the velocity is negative since the rocket turns around and goes back to the ground

• velocity is a vector

• negative y-direction = negative velocity

(6, \infty ) has the same slope, since gravity is the slope

Mass Changing Over Time

Newton’s Second Law of Motion describes Force with a constant mass

• F=\frac{dp}{dt}=m\left(\frac{dv}{dt}\right)=ma

Impulse-Momentum Theorem describes force with a variable mass

• F=\frac{dp}{dt}=v\left(\frac{dm}{dt}\right)

Example 1

Solution

B.)

F = \frac{dp}{dt}

The extinguisher ejects 1.0 kg per second, so its mass changes by -1, at a speed of 6 m/s

F = \left(-1\right)\left(6\right) = -6 N

• To make the extinguisher not acceleratem the person must exert an opposite but equal force

  • F_net = ma

  • F_extinguisher + F_Person = 0

  • -6 + F_person = 0

  • F_person = 6 N

Example 2

Solution

Thrust