(3)

Mass of ball, $m=0.5$ kg

$h_1=40\text{m}$ and $h_2=10\text{m}$

So, velocity of ball before hitting the ground

$v_1=\sqrt{2g{h}_1}=\sqrt{2\times 9.8\times 40}\Rightarrow v_1=28\text{m/s}$

After hitting the ground, let final velocity of the ball be $v_2.$

$\therefore v_2=\sqrt{2g{h}_2}=\sqrt{2\times 9.8\times 10}\Rightarrow v_2=14\text{m/s}$

Change in momentum, $\Delta p=m[v_2-(-v_1\left)\right]$

$=\frac{1}{2}[14-(-28\left)\right]$

$\Delta p=21\text{kg}\Rightarrow \text{Impulse, }\Delta p=21\text{N s}$

(3)

[IMAGE 15]

Impulse is a measure of the degree to which an external force produces a change in momentum of the body. Mathematically

$F=$$\left|\frac{{\displaystyle \Delta \overrightarrow{p}}}{{\displaystyle \Delta t}}\right|$$=\frac{2mv\sin \theta }{t}=\frac{2\left(1\right)\left(1\right)\sin 60°}{0.1}=10\sqrt{3}\text{N}$
 

(1)

[IMAGE 16]-------------------

Given, $\overrightarrow{u}=-u\hat{j}$

$\overrightarrow{v}=u\hat{i}$

Acceleration, $\overrightarrow{a}=\frac{\overrightarrow{v}-\overrightarrow{u}}{t}$

$\overrightarrow{a}=\frac{u\hat{i}+u\hat{j}}{t}$

Direction of acceleration $=\tan \theta =\frac{v}{u}=1$

$\theta =45°$ along north-east, force is also acting towards north-east.

(3)

When bullet covers 0.24 m, speed is $\frac{u}{3}$

Here constant retardation $\left(a\right)$ will be provided to the bullet by frictional force.

Now, applying equation of kinematics,

$\frac{u^2}{9}=u^2-2a(0.24)\Rightarrow \frac{8}{9}{u}^2=a(0.48)$

or     $a=\frac{8u^2}{9\times 0.48}$                              ...(i)

Now, for further motion, let $x$ be the distance when bullet stops.

$0=\frac{u^2}{9}-2ax,$

$\Rightarrow 2ax=\frac{u^2}{9}\Rightarrow a=\frac{u^2}{18x}$                 ...(ii)

From equation (i) and (ii), we get

$\frac{8u^2}{9\times 0.48}=\frac{u^2}{18x}\Rightarrow x=\frac{9\times 0.48}{8\times 18}=3\text{cm}$

Thus, total length = $24\text{cm}+3\text{cm}=27\text{cm}$

(3)

Mass, $m=0.15\mathrm{kg}; h_1=10\mathrm{m}, h_2=10\mathrm{m}; g=10\mathrm{m}/{\mathrm{s}}^2$

The velocity at the time of strike when going downwards.

             $v=-\sqrt{2g{h}_1}$

and when moving up $v\text{'}=\sqrt{2g{h}_2}$

Impulse = change in momentum $mv\text{'}-mv=2mv$ (As $h_1=h_2$)

$=2\times 0.15\times \sqrt{2\times 10\times 10}=4.2\mathrm{kg}$ $\mathrm{m}/\mathrm{s}$

 (3)

[IMAGE 18]------------------------------------------

Change in momentum = Area under $F-t$ graph in that interval = Area of $∆$ABC $-$ Area of rectangle CDEF $+$Area of rectangle FGHI

$=\frac{1}{2}\times 2\times 6-3\times 2+4\times 3=12$ $\text{N}$ $s$