6.1 Instantaneous impulse

1.1 Position of centre of mass for a system of two particles

2.1 Rod
2.2 Semicircular ring
2.3 Semicircular disc
2.4 Solid sphere
2.5 Hemispherical Shell
2.6 Rectangular Plate
2.7 Square Plate
2.8 Circular Plate
2.9 Solid Cone
2.10 Hollow Cone
2.11 Questions
2.12 Centre of mass of a rigid complex bodies

3.1 Motion of mass of the system

5.1 Rocket Propulsion

6.1 Instantaneous impulse

9.1 For direct collision
9.2 For collision in two dimension
9.3 For oblique collision

10.1 Head on elastic collision
10.2 Head on inelastic collision

Question 18. A $2kg$ ball drops on floor from a height of $20m$ and rebounds with $20%$ of the initial speed. Find the impulse received by the ball.

Solution: Velocity $\left( {{{\overrightarrow v }_i}} \right)$ when the ball hits the ground,

$$\begin{equation} \begin{aligned} v_i^2 = {u^2} + 2\overrightarrow g \overrightarrow h \\ v_i^2 = 0 + 2( - 10)({h_f} - {h_i}) \\ v_i^2 = 0 + 2( - 10)(0 - 20) \\ {v_i} = 20m/s{\text{ (downward)}}\quad ...(i) \\\end{aligned} \end{equation} $$

Initial linear momentum, $$\left( {{{\overrightarrow p }_i}} \right) = m{\overrightarrow v _i} = 2( - 20)\widehat j = - 40\widehat j\;N/s\quad ...(ii)$$

Velocity $\left( {{{\overrightarrow v }_f}} \right)$ with which the ball rebounds, $$\begin{equation} \begin{aligned} {v_f} = 0.20\left( {{v_i}} \right) \\ {v_f} = 4\;m/s{\text{ (upwards)}}\quad ...(iii) \\\end{aligned} \end{equation} $$

Final linear momentum $$\left( {{{\overrightarrow p }_f}} \right) = m{\overrightarrow v _f} = 2( + 4)\widehat j = 8\widehat j\;N/s\quad ...(iv)$$

Change in linear momentum $\left( {\Delta \overrightarrow p } \right)$, $$\begin{equation} \begin{aligned} \Delta \overrightarrow p = {\overrightarrow p _f} - {\overrightarrow p _i} \\ \Delta \overrightarrow p = 8\widehat j - ( - 40)\widehat j \\ \Delta \overrightarrow p = 48\widehat {j\;}\;N/s \\\end{aligned} \end{equation} $$

Note: $\widehat {j\;}$ is a unit vector in the vertical direction.

Question 19.An inextensible string is passing over a light frictionless pulley. One end of the string is connected to a block of mass $4m$ which is resting on a horizontal surface. To another end, a plate of mass $m$ is attached and is hanging in the air. A ball of mass $m$ strikes from above with vertical velocity $v_0$ and sticks to the plate. Calculate maximum height attained by a block of mass $4m$.

Solution: Initially the system is at rest.

When the ball of mass $m$ strikes the pan with a velocity $v_0$, it generates tension $T$ in the string.

Let us consider an impulse $I$ due to the tension $T$ for an infinitesimally small time $dt$, $$I = Tdt$$

Momentum before impact $\left( {{{\overrightarrow p }_i}} \right)$, $${\overrightarrow p _i} = - m{v_0}\widehat j\quad ...(ii)$$

After the impact, ball and pan sticks with each other and move with a velocity $v$.

Momentum after impact $\left( {{{\overrightarrow p }_f}} \right)$, $${\overrightarrow p _f} = - 2mv\;\widehat j\quad ...(iii)$$

Impulse for the ball and pan is given by, $$\begin{equation} \begin{aligned} \overrightarrow I = {\overrightarrow p _f} - {\overrightarrow p _i} \\ \overrightarrow I = - 2mv\;\widehat j - \left( { - m{v_0}\;\widehat j} \right) \\ \overrightarrow I = - 2mv\;\widehat j + m{v_0}\;\widehat j\quad ...(iv) \\\end{aligned} \end{equation} $$

Momentum before impact $\left( {{{\overrightarrow p }_i}} \right)$, $${\overrightarrow p _i} = 0\quad ...(v)$$

After the impact, block of mass $4m$ also moves with velocity $v $ as the string is inextensible.

Momentum after impact $\left( {{{\overrightarrow p }_f}} \right)$, $${\overrightarrow p _f} = + 4mv\;\widehat j\quad ...(vi)$$

Impulse for the ball and pan is given by, $$\begin{equation} \begin{aligned} \overrightarrow I = {\overrightarrow p _f} - {\overrightarrow p _i} \\ \overrightarrow I = 4mv\;\widehat j - 0 \\ \overrightarrow I = 4mv\;\widehat j\quad ...(vii) \\\end{aligned} \end{equation} $$

From equation $(iv)$ & $(vii)$, we get, $$\begin{equation} \begin{aligned} 4mv\;\widehat j = - 2mv\;\widehat j + m{v_0}\;\widehat j \\ 6mv\;\widehat j = m{v_0}\;\widehat j \\ v = \frac{{{v_0}}}{6}\quad ...(viii) \\\end{aligned} \end{equation} $$

From FBD we can write, $$\begin{equation} \begin{aligned} T - 2mg = 2ma\quad ...(ix) \\ 4mg - T = 4ma\quad ...(x) \\\end{aligned} \end{equation} $$

Adding the above equation we get, $$\begin{equation} \begin{aligned} 2mg = 6ma \\ a = \frac{g}{3}\quad ...(ix) \\\end{aligned} \end{equation} $$

The equation of motion for the block can be written as, $$\begin{equation} \begin{aligned} {v^2} = {u^2} + 2\overrightarrow g \overrightarrow s \\ 0 = {\left( {\frac{{{v_0}}}{6}} \right)^2} + 2\left( { - \frac{g}{3}} \right)\left( s \right) \\ s = \frac{{v_0^2}}{{24g}} \\\end{aligned} \end{equation} $$