Lecture1 [PHY 117]

 Lecture 1

Date: 28.06.20

Linear Momentum

    Center of Mass: The center of mass of a system of particles is the point that moves as through,

        (a) all the system's mass were concentrated there and

        (b) all external forces were applied there.

The position of the center of mass is now defined

$$X_{CM} = \frac{m_1x_1+m_2x_2}{m_1+m_2}$$

    Many Particles: We can extend this equation to a more general situation in which "n" particles are stung out along the x-axis. Then the total mass is,

$$\begin{align*} &M= m_1+m_2+m_3\cdots +m_n \\ &M_r= m_1r_1+m_2r_2+m_3r_3+\cdots+m_nr_n \end{align*}$$

The expression of the center of mass for many bodies, $$r_{cm}= \frac{m_1r_1+m_2r_2+m_3r_3+\cdots+m_nr_n}{M}$$

For three Dimension: - $$r_{cm}= \frac{1}{M}\sum_{i=1}^n m_ir_i$$

    Momentum: We know that it is harder to get a more massive object moving from rest than a less massive object. This is the concept of inertia.

We call this new concept momentum and it depends on the mass of the object and how fast it is moving.

    (a)The linear momentum of an object is defined as the product of its mass and its velocity. It is measured in $kg.ms^{-1}$. The linear momentum of a particle is a vector quantity $\vec{P}$ that defined as, $\vec{P}=m \vec{v}$ in which $m$ is the mass of the particle and $\vec{v}$ is its velocity.
    (b)Newton expressed his second law of motion in terms of momentum: The time rate of change of the momentum of a particle is equal to the net force acting on the particle and is in the direction of that force. In equation form this becomes, $$\begin{align*} F_{net} & =\frac{d\vec{P}}{dt} \\ & =\frac{d(m\vec{v})}{dt}\\ & = m\vec{a} \end{align*}$$
    (c)Let's extend the definition of linear momentum to a system of particle. Consider a system of n particles, each with its own mass, velocity and linear momentum. The system as a whole has a total linear momentum $\vec{P}$ . which is defined to be the vector sum of the individual particles linear momentum. Thus, $$\begin{align*} \vec{P} & =\vec{P_1}+\vec{P_2}+\vec{P_3}+\cdots+\vec{P_n}\\ & =m_1\vec{v_1}+m_2\vec{v_2}+m_3\vec{v_3}+\cdots+m_n\vec{v_n} \end{align*}$$
So, the linear momentum of a system of particles is equal to the product of the total mass M of the system and the velocity of the center of mass.

    

    Impulse: The momentum of any particle-like body cannot change unless a net external force changes it. For example, we could arrange for the body to collide with a baseball bat. In such a collision or crash the external force on the body is brief, has a large magnitude and suddenly changes the body's momentum. Let the projectile be a ball and the target be a bat. The collision is brief, and the ball experiences a force that is great enough to slow, stop or even reverse its motion. The figure depicts the collision at one instant. We see that in the time interval $dt$ the change in the ball's momentum is $$d\vec{P}=F(t)~dt$$

We can find the net change in the ball's momentum due to the collision if we integrate both sides of the equation from a time $t_i$ just before the collision to a time $t_f$ just after the collision $$\int_{P_i}^{p_f}d\vec{P}=\int_{t_i}^{t_f}F(t)~dt$$

The left side of this equation gives us the change in momentum: $\vec{P_f}-\vec{P_i}=\Delta \vec{P}$

The right side, which is a measure of both the magnitude and the duration of collision force, is called the impulse $\vec{J}$ of the collision: $$\begin{align*} &\vec{J}=\int_{t_i}^{t_f}F(t)~dt \\ &\Delta P_x=J_x \\ &P_{f_x}-P_{i_x}= \int_{t_i}^{t_f}F_x~dt \end{align*}$$

If we have a function for we can evaluate (and thus the change in momentum) by integrating. In many situations, we do not know how the force varies with time but we do know the average magnitude $F_{avg}$ of the force and the duration $\Delta t$ of the collision. Then we can write the magnitude of impulse, $J=F_{avg}~\Delta t$

Impulse is the area under $F(t)$ curve.