0801 Momentum and Impulse

Newton's Second Law in Terms of Momentum

Consider a particle of constant mass $m$ . Because $\vec a = \dfrac{d\vec v}{dt}$ ,

$\begin{aligned} \sum \vec F = m \cdot \frac{d\vec v}{dt} = \frac{d}{dt}(m\vec v) \end{aligned}$

The net force $\sum \vec F$ acting on a particle equals the time rate of change of the product of the particle's mass and velocity. This product is called momentum, of the particle:

$\begin{aligned} \vec p &= m\vec v\\ \sum \vec F &= \frac{d\vec p}{dt} \end{aligned}$

The Impulse-Momentum Theorem

The impulse of a net force, denoted by $\vec J$ , is defined to be the product of the net force and the time interval.

$\begin{aligned} \vec J &= \sum \vec F(t_2-t_1) = \sum \vec F \Delta t &\text{(Eq.1)} \end{aligned}$

$\vec F$ is the constant net force.
$t_2-t_1$ is the time interval over which the net force acts.
Impulse is a vector quantity and has the same direction as the net force $\sum \vec F$ .

If the net force $\sum \vec F$ is constant, then $\frac{d\vec p}{dt}$ is also constant. In that case, $\frac{d\vec p}{dt}$ is equal to the total change in momentum $\vec p_1 - \vec p_2$ during the time interval $t_2-t_1$ .

$\begin{aligned} \sum \vec F &= \frac{\vec p_1 - \vec p_2}{t_2-t_1}\\ \sum \vec F (t_2-t_1)&= \vec p_1 - \vec p_2 &\text{multiplying }t_2 -t_1\\ \vec J &= \vec p_1 - \vec p_2 = \Delta p &\text{(Eq.2)} \end{aligned}$

The Impulse-Momentum Theorem: The impulse of the net force on a particle during a time interval equals to the change in momentum of that particle during that interval.
It also holds when forces are not constant. To see this, we integrate both sides of Newton's second law $\sum \vec F = \frac{d\vec p}{dt}$ over time between $t_1$ and $t_2$

$\begin{aligned} \int_{t_1}^{t_2} \sum \vec F dt = \int_{t_1}^{t_2} \frac{d\vec p}{dt} dt = \int_{p_1}^{p_2} d\vec p = \vec p_2 - \vec p_1 \end{aligned}$

We see from Eq.2 that integral on the left is the impulse of the net force:

$\begin{aligned} \vec J = \int_{t_1}^{t_2} \sum \vec F dt \end{aligned}$

The meaning of the area under a graph of $\sum F_x$ versus $t$ .
Graph a Graph b

Exercises

8.8 Force of a Baseball Swing. A baseball has mass 0.148 kg.
(a) If the velocity of a pitched ball has a magnitude of 43.5 m/s and the batted ball's velocity is 60.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
(b) If the ball remains in contact with the bat for 2.40 ms, find the magnitude of the average force applied by the bat.

Solution

a. $P = 15.3 kg\cdot m/s$
b. $F = 6382.5N$

8.11 At time $t = 0$ a $2150 kg$ rocket in outer space fires an engine that exerts an increasing force on it in the $+x$ direction. This force obeys the equation $F_x = At^2$ , where $t$ is time, and has a magnitude of $781.25 N$ when $t = 1.25 s$ .
a. Find the SI value of the constant $A$ , including its units.

Solution

$\begin{aligned} F_x &= A t^2 \\ A &= \frac{F_x}{t^2} = \frac{781.25}{1.25^2} = 500N/s^2 \end{aligned}$

b. What impulse does the engine exert on the rocket during the $1.50 s$ interval starting $2.00 s$ after the engine is fired?

Solution

$\begin{aligned} J &= \int_{2.00}^{2.00 + 1.5} F_x dt\\ &= \int_{2.00}^{3.5} 500t^2 dt\\ &= \frac{500}{3} t^3 \mid_{2.00}^{3.5} = 5812.5 N\cdot s \end{aligned}$

c. By how much does the rocket's velocity change during this interval? Assume constant mass.

Solution

The change of momentum $\Delta p$ is $J = 5812.5N\cdot s$ .

$\begin{aligned} \Delta p &= m \cdot \Delta v \\ \To \Delta v &= \frac{\Delta p}{m} = \frac{5812.5}{2150} \approx 2.7 m/s \end{aligned}$

(8.14) Starting $t=0s$ , a horizontal net force $\vec F = (0.275 N/s) \hat i + (-0.460N/s^2)t^2 \hat j$ is applied to a bos that has an initial momentum $p=(-3.10kg \cdot m/s)\hat i + (3.85kg \cdot m/s)\hat j$ . What is the momentum of the box at $t = 2.15 s$ ?

Solution

$p_x,p_y = -2.46, 2.33 kg \cdot m/s$

(8.16) A 64.5 kg astronaut is doing a repair in space on the orbiting space station. She throws a 2.35 kg tool away from her at 3.50 m/s relative to the space station. Find the speed and direction she begin to move.

Solution

Identify: Conservation of Momentum
Setup: Let the mass of the astronaut be $m_1$ , and the mass of the tool be $m_2$ . Without external force, the total momentum is zero before and after she throws the tool. Let the direction of motion of the astronaut be positive. $v_2=-3.5m/s$

$\begin{aligned} 0 &= m_1 v_1 + m_2 v_2\\ \To & v_1 = -\frac{m_2 v_2}{m_1}\\ &= -\frac{2.35 \times (-3.5)}{64.5} \approx 0.128m/s \end{aligned}$

She will move opposite to the direction in which she throws the tool.

(8.17) The expanding gases that leave the muzzle of a rifle also contribute to the recoil. A $.30$ -caliber bullet has mass 0.00720 kg and a speed of 601 m/s relative to the muzzle when fired from a rifle that has mass 2.80 kg. The loosely held rifle recoils at a speed of 1.85 m/s relative to the earth. Find the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle.

Solution

Identify: The rifle, the bullet and the propellant gases are initially at rest, so the total momentum is zero. After the short is fired, the rifle, the bullet and the propellant all gain momentum.
Set up: Let $+x$ direction be the bullet's direction.

$\begin{aligned} p_1 &= 0 = p_2 = p_r + p_b + p_g\\ p_r &= m_r v_r = 2.80 \times (-1.85)\\ p_b &= m_b v_b = 0.0072 \times 601\\ p_g &= -p_r - p_b = 0.8528 N/s \end{aligned}$

(8.27) Two ice skaters, Daniel (mass 65.0 kg) and Rebecca (mass 45.0 kg), are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 13.0 m/s before she collides with him. After the collision, Rebecca has a velocity of magnitude 8.00 m/s at an angle of $53.1 \degree$ from her initial direction. Both skaters move on the frictionless, horizontal surface of the rink.
(a) What are the magnitude and direction of Daniel's velocity after the collision?
(b) What is the change in total kinetic energy of the two skaters as a result of the collision?

Solution

Todo

(8.35) In July 2005, NASA's "Deep Impact" mission crashed a $372 kg$ probe directly onto the surface of the comet Tempel 1, hitting the surface at $37000 km/h$ relative to the surface. The original speed of the comet at that time was about $40000 km/h$ , and its mass was estimated to be in the range $(0.10-2.5) \times 10^{14}kg$ . Use the smallest value of the estimated mass.
(a) What change in the comet's velocity did this collision produce?

Solution

$|\Delta v| = 1.4 \times 10^{-6} km/h$

(b) Suppose this comet were to hit the earth at $40000 km/h$ and fuse with it. By how much would it change our planet's velocity? (The mass of the earth is $5.97 \times 10^{24} kg$ .)

Solution

$|\Delta v| = 6.7\times 10^{-8} km/h$
Todo

(Quiz0314) A bullet $m_1 = 15g$ with intial velocity of $500m/s$ hits a wood block $m_2 = 0.8kg$ at the edge of a table with height of $0.8m$ and stays inside the block after collision. Find the final velocity of the wood block before it hits the ground.
Graph

Solution

Stage 1, after the bullet hits the block and before the block start to fall. By conservation of momentum

$\begin{aligned} m_1v_0 &= (m_1 + m_2) v_2\\ v_2 &= \frac{m_1v_0}{m_1+ m_2} \\ &= \frac{0.015 \cdot 500}{0.815} = 9.2m/s \end{aligned}$

Stage 2, from the edge of the table to the ground. Accoring to Kinematic Equations

$\begin{aligned} v_{yf}^2 &= v_{y0}^2 + 2ay\\ v_{yf} &= \sqrt {v_{y0}^2 + 2ay} \\ &= \sqrt{0 + 2\cdot 9.8\cdot 0.8 } = 3.96m^2\\ v_f & = \sqrt{v_{xf}^2 + v_{yf}^2}\\ &= \sqrt{9.2^2 + 3.96^2} = 10m/s \end{aligned}$