$$\eqalign{ & \vec J = m\left( {{{\vec v}_f} - {{\vec v}_i}} \right) \cr & = 2\,mv\frac{{\sin \theta }}{2} = 2\,mv\,\sin \frac{\pi }{n}. \cr} $$

Apply 2^nd law of motion i.e., rate of change of momentum is equal to force applied. The vector $$OA$$  represents the momentum of the wall, before the collision and the vector $$OB$$  that after the collision. The vector $$AB$$  represents the change in momentum of the ball $$\Delta P.$$
As, the magnitude of $$OA$$  and $$OB$$  are equal the components of $$OA$$  and $$OB$$  along the wall are equal and in the same direction, while those perpendicular to the wall are equal and opposite. Thus, the change in momentum is only due to the change in direction of the perpendicular components.

Hence, $$\Delta P = OA\sin {60^ \circ } - \left( { - OB\sin {{60}^ \circ }} \right)$$
$$\eqalign{ & = mv\sin {60^ \circ } + mv\sin {60^ \circ } \cr & = 2mv\sin {60^ \circ } \cr & = 2 \times 3 \times 100 \times \frac{{\sqrt 3 }}{2} \cr & = 300\sqrt 3 \,kg - m/s \cr} $$
The force exerted on the wall
$$F = \frac{{\Delta p}}{{\Delta t}} = \frac{{300\sqrt 3 }}{{0.2}} = 1500\sqrt 3 N$$

$$\eqalign{ & {\text{Here}}\,\,u = 10\,m{s^{ - 1}},v = - 2\,m{s^{ - 1}},t = 4\,s,a = ? \cr & {\text{Using}}\,a = \frac{{v - u}}{t} = \frac{{ - 2 - 10}}{4} = - 3\,m/{s^2} \cr & \therefore {\text{Force,}}\,F = ma = 10 \times \left( { - 3} \right) = - 30\,N \cr} $$

As we know, $$\left| {{\text{impulse}}} \right| = \left| {{\text{change}}\,{\text{in}}\,{\text{momentum}}} \right|$$
$$ = \left| {{p_2} - {p_1}} \right| = \left| {0 - m{v_1}} \right| = \left| {0 - 3 \times 2} \right| = 6Ns$$

Impulse of a force can be calculated as the product of large force applied to the small time to which force act.
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,\,F = \frac{{dp}}{{dt}} \cr & \Rightarrow F \cdot dt = dp \cr & \Rightarrow {\text{impulse}} = {p_2} = {p_1} \cr} $$
Impulse of a force, which is the product of average force during impact and the time for which the impact lasts, is measured by the total change in linear momentum produced during the impact.
Here, $${p_1} = m{v_1},{p_2} = m{v_2}$$
Impulse, $$I = m{v_2} - m{v_1} = m\left( {{v_2} - {v_1}} \right)$$

During each collision
Initial velocity $$ = \frac{2}{2} = 1\,m{s^{ - 1}}$$
Final velocity $$ = - \frac{2}{2} = - 1\,m{s^{ - 1}}$$
Impulse = Change in momentum
$$ = m\left| {{v_2} - {v_1}} \right| = 0.4 \times 2 = 0.8\,Ns$$

The area under $$F$$-$$t$$ graph gives change in momentum.
For 0 to $$2s,$$ $$\Delta {p_1} = \frac{1}{2} \times 2 \times 6$$
$$ = 6\,kg{\text{-}}m/s$$
For 2 to $$4s,$$ $$\Delta {p_2} = 2 \times - 3 = - 6\,kg{\text{-}}m/s$$
For 4 to $$8s,$$ $$\Delta {p_3} = 4 \times 3 = 12\,kg{\text{-}}m/s$$
So, total change in momentum for 0 to $$8s$$
$$\eqalign{ & \Delta {p_{{\text{net}}}} = \Delta {p_1} + \Delta {p_2} + \Delta {p_3} \cr & = \left( { + 6 - 6 + 12} \right) \cr & = 12\,kg{\text{-}}m/s \cr & = 12\,N{\text{-}}s \cr} $$
NOTE
Graphs on negative axis gives - ve momentum.

The vector $$OA$$ represents the momentum of the object before the collision, and the vector $$OB$$  that after the collision. The vector $$AB$$  represents the change in momentum of the object $$\Delta p.$$
As the magnitudes of $$OA$$ and $$OB$$  are equal, the components of $$OA$$ and $$OB$$  along the wall are equal and in the same direction, while those perpendicular to the wall are equal and opposite. Thus, the change in momentum is only due to the change in direction of the perpendicular components.
Hence, $$\Delta p = OB\sin {30^ \circ } - \left( { - OA\sin {{30}^ \circ }} \right)$$
$$\eqalign{ & = mv\sin {30^ \circ } - \left( { - mv\sin {{30}^ \circ }} \right) \cr & = 2mv\sin {30^ \circ } \cr} $$
Its time rate will appear in the form of average force acting on the wall.
$$\eqalign{ & \therefore F \times t = 2mv\sin {30^ \circ } \cr & {\text{or}}\,F = \frac{{2mv\sin {{30}^ \circ }}}{t} \cr & {\text{Given,}}\,m = 0.5\,kg,v = 12\,m/s,t = 0.25\,s \cr & \theta = {30^ \circ } \cr & {\text{Hence,}}\,F = \frac{{2 \times 0.5 \times 12\sin {{30}^ \circ }}}{{0.25}} \cr & = 24\,N \cr} $$

As we know that, impulse is imparted due to change in perpendicular components of momentum of ball.
$$\eqalign{ & J = \Delta p = m{v_f} - m{v_i} \cr & = mv\cos {60^ \circ } - \left( { - mv\cos {{60}^ \circ }} \right) \cr & = 2mv\cos {60^ \circ } \cr & = 2mv \times \frac{1}{2} = mv \cr} $$

According to Newton's 2^nd law, force applied on an object is equal to rate of change of momentum.
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,F = \frac{{dp}}{{dt}} \cr & {\text{or}}\,F = m\frac{{dv}}{{dt}}\,.......\left( {\text{i}} \right) \cr} $$
$${\text{Given,}}\,\,m = 3\;kg,t = 3\;s,F = \left( {6{t^2}\hat i + 4t\hat j} \right)N$$
Substituting these values in Eq. (i), we get
$$\eqalign{ & \left( {6{t^2}\hat i + 4t\hat j} \right) = 3\frac{{dv}}{{dt}} \cr & {\text{or}}\,\,dv = \frac{1}{3}\left( {6{t^2}\hat i + 4\hat j} \right)dt \cr} $$
Now, taking integration of both sides, we get
$$\eqalign{ & \int d v = \int_0^t {\frac{1}{3}} \left( {6{t^2}\hat i + 4t\hat j} \right)dt \cr & v = \frac{1}{3}\int_0^t {\left( {6{t^2}\hat i + 4t\hat j} \right)} dt \cr & {\text{but}}\,\,t = 3\;s\,\,\left( {{\text{given}}} \right) \cr & \therefore v = \frac{1}{3}\int_0^3 {\left( {6{t^2}\hat i + 4t\hat j} \right)} dt \cr & {\text{or}}\,\,v = \frac{1}{3}\left[ {\frac{{6{t^3}}}{3}\hat i + \frac{{4{t^2}}}{2}\hat j} \right]_0^3 \cr & {\text{or}}\,\,v = \frac{1}{3}\left[ {2{{\left( 3 \right)}^3}\hat i + 2{{\left( 3 \right)}^2}\hat j} \right] \cr & {\text{or}}\,\,v = \frac{1}{3}\left[ {54\hat i + 18\hat j} \right] \cr & {\text{or}}\,\,v = 18\hat i + 6\widehat j \cr} $$