$m_A=200 \mathrm{kg}$

$m_B=300 \mathrm{kg}$

Blocks A and B are connected by a light cable on a frictionless pulley (also massless).

The system is released from state of rest.

Find the velocity of block A after it has moved two meters.

$\mu_k$: coefficient of friction between A and table = 0.25

Draw the free body diagrams of bodies A and B.

The string is pulling the body A with a force T.

If a force T is being applied to pull body A the string will pull body B with the same force T.

Forces which are acting:

String force

Weight

The contact force, will consist of a normal reaction.

Friction force

$a_y=0$

$N_A=m_ag$

$f=\mu_k N_A$

$f=\mu_k m_A g =0.25 \times 200 \times 9.8 =490 \mathrm{N}$

Work Energy Principle on block

$\Delta K+\Delta V$ = $W_{\text{other force}}$

$\text { state: } 1 \rightarrow \text { rest }$

$\text { state 2} \rightarrow {v}$

$K_2=\frac{1}{2} m_A v^2$

$K_1=0$

This is for change in kinetic energy, $\Delta K$

The block is moving in a horizontal plane due to gravitational potential energy, $V_1=V_2$.

The block is moving in a horizontal plane due to gravitational potential energy, $V_1=V_2$.

We call this as the reference state.

$K_2=\frac{1}{2} m_A v^2$

$K_1=0$

This is for change in kinetic energy, $\Delta K$

$V_1=V_2$ $\Rightarrow \Delta V=0$

$W_{\text{other force}}:$ $\left(T-\mu_k m_A g\right).s$

$s = 2 m$

$\frac{1}{2} m_A v^2=(T-490) 2$

Now, there are 2 unknowns V and T.

Block:2

$\Delta K+\Delta V=\text W_{ other force. }$

$\Delta K=\left(\frac{1}{2} m_B v^2-0\right)$

$\Delta V=\left(V_2-V_1\right)$

$V_1=0$

$V_2 =-mgs$

$\Delta V =-2 m g$

T is acting upwards

Work done by T on block B: - T(2)

$\frac{1}{2} m_B v^2-m_B g(2)=-2T$

(1) + (2)

$\frac{1}{2} m_A v^2+\frac{1}{2} m_B v^2-m_B g(2)=-2 \times 490$

$v=4.427 \mathrm{~m} / \mathrm{s}$

This problem illustrates that if we consider A and B, then the tension which acts on these bodies separately, but because the work done by tension on body A = -W by tension on body B.

The work done by T cancels out.

$\Delta K + \Delta V$ = $W_{other force}$

Also referred to as principle of conservation of mechanical energy.

This has been derived from Newton's second law.

Rest: We calculate velocity or displacements, with respect to an inertial frame.

$\vec{P}=$Linear Momentum

Define: $\vec{I}=$ $\int_{t_1}^{t_2} \vec{F} d t$

Impulse of a Force, $\vec{F}$

from $(t_1$ to ${t_2})$

Impulse of a force $\vec{F}$ during time internal from $(t_1$ to ${t_2})$ $\rightarrow$$\vec{I}$

$\vec{I}=$ $\int_{t_1}^{t_2} \vec{F} d t$

If $\vec{F}$ is constant, $\vec{I}=\vec{F}\left(t_2-t_1\right)$.

$\vec{F}=\frac{d \vec{P}}{d t}$ (rate of change of linear momentum $\vec{P}$).

$\vec{F}dt=d\vec{P}$.

$\int_{t_1}^{t_2}\vec{F}dt=\int_{P_1}^{P_2}d\vec{P}=\left(\overrightarrow{P_2}-\overrightarrow{P_1}\right) =\Delta \vec{p}$

$t_1 \rightarrow \overrightarrow{P_1}$

$t_2 \rightarrow \overrightarrow{P_1}$

Impulse of force $\vec{F}$ a particle from $t_1$ to ${t_2}$

$\vec{I}=\Delta\vec{P}$

The change in linear momentum of a particle is given by the impulse of the forces acting on the particle.

Principle of impulse is useful if the force is a function of time.

$\vec{F}=\vec{F}(t)$

Some salient features

a) Impulse is a vector quantity. Vector equation can write as scalar components.

$I_x=P_{2, x}-P_{1, x}=m\left(v_{2, x}-v_{ 1 ,x})\right.$

$I_y=P_{2, y}-P_{1, y}=m\left(v_{2,y}-v_{ 1 ,y})\right.$

The dimension of impulse is force: M $\frac{L}{T^2}.T$

= $\frac{ML}{T}$

SI units: N.sec

For a single particle:

Impulse methods are useful if $\vec{F}$ is a function of times.

Impulse give us change in momentum.

$\vec{I}=m(\vec{v_2}-\vec{v_1})$

$m \overrightarrow {v_2 } =m \overrightarrow{v_1}+\vec{I}$

$m \overrightarrow{v_1}$ = Initial momentum

$m \overrightarrow {v_2 }$ = Final momentum

$\vec{I}$ = Impulse

Area under the F-t curve given impulse.

$m v_1+\text { area of triangle }=m v_2$

$\vec{I}=\Delta\vec{P}$

If $\vec{I}=O,\Delta\vec{P}=O$

m$\vec{v_1}=m\vec{v_2}$

- Conservation of Linear Momentum.

$\vec{I_{\text {inst. }}}=\int_{t_1}^{t_1+\epsilon} \vec{F} d t$

limit $\epsilon\rightarrow{0}$

$\vec{F}\rightarrow$ very large $\rightarrow{\infty}$

$F_{av.}(\Delta{t})\rightarrow$${finite}$

$F_{av.}$= Large

$(\Delta{t})$= Very small

eg. Roger Federer hitting a tennis ball.

The time of contact between the ball and the racket is very small.

Contact force ${\rightarrow}$ large.

Similar example: Virat Kohli hits the cricket ball with bat.

a) change direction

b) speed of particle is also changed.

When an impulsive force acts on a particle other finite forces may also be acting during this interval.

Because Impulsive Force is very large.

($t_1$ to $t_1 +\epsilon$) $\longrightarrow$

During the time period ${\epsilon}$, we neglect the effect of the other finite forces.

Draw force versus time.

During time period from $t_1$ to $t_1 +\epsilon$, only the effect of the impulsive force is counted.

Average contact force $\overrightarrow{F_{a v}}$.

Time of contact: $\Delta{t}$.

$\vec{F}_{\text {av. }} \Delta t=\Delta \vec{p}$

If know $\vec{P_1}$ and $\overrightarrow{P_2}$, we can find $\vec{F}_{av. }$

$\vec{P_1}=m\vec{v_{1}}$

$\vec{P_2}=m\vec{v_{2}}$

Let us see how principle of momentum can be used when we have more than one particle.

Conservation of momentum

Collision Problems

Post collision

The second type of problem is when we have one body which is moving and it suddenly breaks into two or more parts.

Case of two particles

We treat both the particles as one system.

If no external force acts on the two particles then the momentum of both these particles together as a system is conserved.

FBD of A: $\vec{f_{A B}}$, the force exerted on particle A by particle B.

FBD of B: $\vec{f_{BA}}$, Force exerted on B by A.

When we add up the 2 systems

Newtons third law: $\vec{f_{A B}}=-\vec{f_{B A}}$.

Impulse momentum principle for particle A

$\int \vec{F_A} d t+\int \vec{f}_{A B} d t=\Delta \vec{P_A}$

Impulse momentum principle for particle B

$\int \vec{F_B} d t+\int \vec{f}_{BA} d t=\Delta \vec{P_B}$

Add: $\int\vec{(F_A}+\vec{F_B)}dt+O=\Delta\vec{P_A}+\Delta\vec{P_B}$

If $\vec{F_A}$ and $\vec{F_B}=O$

$\Delta\vec{P_A}+\Delta\vec{P_B}=O$

${(m_A\vec{v_A}}+{m_B\vec{v_B})}_1={(m_A\vec{v_A}}+{m_B\vec{v_B})}_2$

During the collision period collision forces >> other finite forces

For the period of collision if both particles treated as a system, momentum of the system is conserved.

Angular Momentum or Moment of Momentum

The position vector of the particle: $\vec{OP}= \vec{r}$.

Angular Momentum (Momentum of Momentum) of the particle about point 0

$\equiv\vec{r}\times{m \vec{v}}$ = $\vec{H_o}$

${m \vec{v}}$= Linear Momentum

$\vec{(H_o)_1}=\vec{r_1}\times{m\vec{v_1}}$

$\vec{(H_o)_2}=\vec{r_2}\times{m\vec{v_2}}$

$\vec{H_o}$: angular momentum is a vector.

$\vec{H_o}$ = $\vec{r}\times{m\vec{v}}$

Angular momentum is perpendicular to $\vec{r}$ and it is perpendicular to $\vec{v}$.

$\vec{H_0}=\vec{r} \times m \vec{v}$

$\frac{d \vec{H_0}}{d t}=\frac{d}{d t}(\vec{r} \times m \vec{v})$

$\frac{d \vec{r}}{d t} \times m \vec{v}+\vec{r} \times {m}\frac{d \vec{v}}{d t}$

$\vec{v}=\frac{d \vec{r}}{d t}$

$\vec{r} \times{m} \frac{d \vec{v}}{d t}=\vec{r}\times{m}\vec{a}$

$\vec{V}\times{m\vec{V}}=0$

$\frac{d \vec{H_0}}{d t}=\vec{r} \times m\vec{a}$

$m\vec{a}=\vec{F}_{\text{ext.on particle}}=\vec{F}$

$\frac{d \vec{H_0}}{d t}=\vec{r} \times \vec{F}$

$\vec{r} \times \vec{F}$= Moment of force about point O.

$\frac{d \overrightarrow{H_0}}{d t}=\overrightarrow{m}_0$

If $\vec{m_0}=0$

$\frac{d \overrightarrow{H_0}}{d t}=0$

$\vec{H_0}$=Constant

$\vec{(H_0)_1}=\vec{(H_0)_2}$

$\Rightarrow$ If $\vec{m_0}=0$

then $\vec{H_0}$ is conserved.

Useful Planetary motion.

Gravitational force =$\frac{-Gm_1m_2}{r^2}$

It acts towards the center of the planet.

For the motion of satellite about planet $\vec{H_0}$ will be constant.

If particle is under the motion of a force which always point towards fixed point O,

$(\vec{r} \times m \vec{v})_1=(\vec{r} \times m \vec{v})_2$

where, $\vec{r}$ is the position vector with respect to O.