Solved Problems

(a) If a force $F$ is applied to a particle for a time interval $\Delta t$, it causes the momentum $P$ of the particle to change. According to the original form of Newton's second law, the rate of change of the momentum of a particle is equal to the force acting on it.

In this case, since the particle rebounds back with no loss of speed, the change in its momentum is calculated as follows: \begin{align*} \Delta \vec{P} &=P_{rebound}-P_{impact} \\\\ &=m(+v)-(m(-v))\\\\ &=+2m\end{align*} (we have considered the positive direction of the coordinate to be to the right). Therefore, we have  \begin{align*} F_{max}&=\frac{\Delta P}{\Delta t}\\\\ &=\frac{2mV}{\Delta t} \\\\ &=\frac{2\times 0.1\times 5}{10\times {10}^{-3}} \\\\ &=100\,\rm N \end{align*}  

(b) Using Newton's second law, $a_x=F/m$, the ball has acceleration only in the time interval $10\,\rm ms$. 

(c) To draw the velocity vs. time graph, use the following kinematic equation for motions with constant acceleration, $v=v_0+a\Delta t$.

In such cases, the different sections of the velocity vs. time graph are a straight line.

In this problem, the first section ($\rm{I}$) has$a_x=0$, so $v=v_0=-5\,\rm m/s$ which corresponds to a line parallel to the time axis. 

In the second section ($\rm{II}$), $a_x=1000\,\rm m/s^2$, so \[v=v_0+a\Delta t \to v=-5+1000\,t\] Which is a diagonal line with a slope $1000$. 

The third section ($\rm{III}$) is the same as the first.

(a) Since the two cars stick together after the collision, we have a completely inelastic collision. In this type of collision, the momentum of the system is conserved but the kinetic energy is not. Therefore, applying the conservation of the momentum before and after the collision, we have \begin{gather*} \vec P_i=\vec P_f \\\\ m_c{\vec{v}}_c+m_t\underbrace{{\vec{v}}_t}_{0}=\left(m_c+m_t\right){\vec{v}}_f \\\\ \Rightarrow {\vec{v}}_f=\frac{m_c}{m_c+m_t}{\vec{v}}_c \end{gather*} Substituting the given numerical values, gives \[ {\vec{v}}_f=\frac{1200}{1200+2000}\times (12)=5.63\,\rm m/s\] 
(b) The impulse ($\vec{J}$) is the product of the average force and the time interval during which this force acts on an object. It is equal to the change in momentum: $\vec{J}={\vec{F}}_{ave}\Delta t=\Delta \vec{P}$. Therefore, \begin{align*} {\vec{J}}_{car}&=\Delta {\vec{P}}_{car} \\ &=m_{car}({\vec{v}}_f-{\vec{v}}_i) \\ &=1200(5.63-15) \\&=-1.13\times {10}^4\,\rm kg\cdot m/s\end{align*}
(c) By definition of the average acceleration, we have \begin{align*} {\overline{a}}_{ave}&=\frac{\Delta \vec{v}}{\Delta t} \\\\ &=\frac{v_f-v_i}{\Delta t} \\\\ &=\frac{5.63-15}{0.5} \\\\ &=-18.\,\rm m/s^2\end{align*} Now apply Newton's second law of motion to the car and find its average force \begin{align*} F_{ave}&=ma_{ave} \\ &=1200\times (-18.8) \\ &=-22.5\,\rm kN \end{align*} 
(d) Find the following ratio \begin{align*} e&=\frac{K_i-K_f}{K_i} \\\\ &=\frac{\frac{1}{2}m_cv^2_c-\frac{1}{2}(m_c+m_t)v^2_f}{\frac{1}{2}m_cv^2_c} \\\\ &=\frac{1200\times {15}^2-3200\times (5.63)^2}{1200\times {15}^2} \\\\ \Rightarrow \quad e&=\boxed{0.625}\end{align*}  

(a) There is no external force (such as friction) in the $x$ direction, so from the principle of the conservation of momentum, we have:
\begin{gather*} {\vec{P}}_i={\vec{P}}_f \\\\ (m_1+m_2){\vec{V}}_i=m_1{\vec{V}}_{1f}+m_2{\vec{V}}_{2f} \\\\ (2+0.5)(+2)=0.5\times (-2)+2V_{2f} \\\\ \Rightarrow V_{2f}=+3\,\rm m/s \end{gather*} The positive sign indicates that $2\,\rm kg$ block is moving to the right (since we have considered the positive direction to be to the right).

(b) Using the conservation of mechanical energy $E_1=E_2$, we obtain 
\[\frac{1}{2}k\Delta x^2+\frac{1}{2}(m_1+m_2)V^2_i=\frac{1}{2}m_1V^2_{1f}+\frac{1}{2}m_2V^2_f\] Where $\frac{1}{2}kx^2$ represents the elastic potential energy of the spring. Therefore, solving for $\Delta x^2$, and substituting the given values, yields: \begin{gather*} \Delta x^2=\frac{1}{k}\left(m_1V^2_{1f}+m_2V^2_{2f}-\left(m_1+m_2\right)V^2_i\right) \\\\ \Delta x^2=\frac{1}{25000}\left(0.5\times {\left(-2\right)}^2+2\times 3^2-\left(2.5\right)\times 2^2\right) \\\\ \Delta x=\sqrt{\frac{10}{25000}}=\sqrt{4\times {10}^{-4}}=0.02\,\rm m\end{gather*} In the last equality, the square root of both sides is taken.

To reach the initial height, the velocity just before and after the bounce must have equal magnitude but opposite direction (because of conservation of mechanical energy).

According to the conservation of mechanical energy, the initial mechanical energy is equal to the final mechanical energy. We can express this as: \begin{gather*} E_i=E_f \\\\ \Rightarrow  mgh=\frac{1}{2}mv^2\to v=\sqrt{2gh} \end{gather*}

When a net force acts on an object, the impulse of this force is equal to the change in the momentum of the object that is $\Delta \vec{P}=\Delta \vec{F}\cdot \Delta t$. Substituting the numerical values into this gives \begin{align*} \Delta \vec{P}&=m({\vec{v}}_2-{\vec{v}}_1) \\\\ &=m (v\hat{j}-v(-\hat{j})) \\\\ &=2mv\ \hat{j} \end{align*} Here, ($\vec{v}_1$) represents the velocity just before the bounce, and ($\vec{v}_2$) represents the velocity just after the bounce. Substituting $\sqrt{2gh}$ for $v$ in the above equation, we get \begin{align*} \Delta P &=2m\sqrt{2gh} \\\\ &=2(0.250)\sqrt{2(9.8)(1.8)} \\\\ &=2.97\,\rm kg\cdot m/s \end{align*}

In this problem, releasing the spring can be viewed as an explosion, therefore momentum and mechanical energy must be conserved. 
\[{\vec{P}}_{initial}={\vec{P}}_{final}\] 
\[0=M\vec{v}+m{\vec{v}}_1\to v_1=-2v\] 
Initially the system, due to the spring, has potential energy. After releasing the spring potential energy converts to the kinetic energy, so in final we have
\[U_{initial}=K_{final}=\frac{1}{2}Mv^2+\frac{1}{2}mv^2_1\] 
\[=\frac{1}{2}\left(2m\right)v^2+\frac{1}{2}m{\left(-2v\right)}^2=3mv^2\] 
\[\Rightarrow U_{initial}=3mv^2\] 
 

We can treat this as two separate events 
(i) Bullet colliding with the block: to find the speed of the block, we’ll use the conservation of momentum.
(ii) Block (and bullet) sliding with friction: To determine the distance, we’ll use the conservation of energy or the work-energy theorem.

Let’s start with the conservation of momentum. The combined speed of the block and bullet after the collision can be calculated as follows: \begin{align*} P_i&=P_f\\\\ m_{bul}\,v_{i,bul}&=(m_{bul}+M_{blo})V_f \\\\ \Rightarrow V_f&=\frac{m_{bul}}{m_{bul}+M_{blo}}v_{i,bul}\\\\&=\left(\frac{0.0105\,{\rm kg}}{0.0105+10.5}\right)750\\\\ &=0.75\quad {\rm m/s}\end{align*} where $W_f$ is the work done by the friction force. The remaining part is a kind of work-energy theorem problem. Apply this principle to find the desired distance. 
\begin{align*} K_2-K_1&=W_{net} \\\\ 0-\frac{1}{2}(m_{bul}+M_{blo})V^2_f&=W_f\\\\&=-f_kd\\\\&=-{\mu }_k(m_{bul}+M_{blo})gd \\\\ \Rightarrow V_f^2 &=2{\mu }_kgd \end{align*} Solving the above equation for $d$, and substituting the numerical values, we have
\begin{align*} d&=\frac{V^2_f}{2{\mu }_kg}\\\\&=\frac{(0.75)^2}{2\times 0.22\times 9.8}\\\\&=0.13\quad {\rm m}\end{align*} Because, we want the system to come to a stop, so $K_1=0$.

In the above expressions, $W_f$ is the work done by the friction force $f_k$.

If you want to more about static and kinetic coefficient friction force, refer to the following page: 
Coefficient of friction problems and solutions

(a) An elastic collision conserves both momentum and energy: 
\begin{gather*} m{\vec{V}}_i=m{\vec{V}}_8+m{\vec{V}}_c \\\\ \frac{1}{2}mV^2_i=\frac{1}{2}mV^2_8+\frac{1}{2}mV^2_c \\\\ \Rightarrow V^2_i=V^2_8+V^2_c \end{gather*} Where $\vec{V_i}$ is the initial velocity of the cue ball.

From the geometry above, we can obtain \[{\vec{V}}_i={\vec{V}}_8+{\vec{V}}_c\] Squaring both sides of the above relation and using the result of the conservation of kinetic energy, we get
\begin{gather*} {\vec{V}}^2_i=({\vec{V}}_8+{\vec{V}}_c) \cdot ({\vec{V}}_8+{\vec{V}}_c) \\\\ \Rightarrow V^2_i=V^2_8+V^2_c+2{\vec{V}}_8 \cdot {\vec{V}}_c \end{gather*} Substituting the values gives: \begin{align*} V^2_i&=V^2_8+V^2_c+2V_8V_c\,\cos (\pi -\varphi) \\\\&=\underbrace{V^2_8+V^2_c}_{V^2_i}-2V_8V_c \cos \varphi \end{align*} Equating both sides, gives \[\cos \varphi=0\] Therefore, we conclude that $\phi =90^\circ$, i.e., the angle of cue ball with respect to $\vec{V_i}$ must be $60^\circ$.

(b) Now we can use the conservation of the momentum in $x$ and $y$ directions, separately.

 \begin{gather*} P_{ix}=mV_i \\\\ P_{fx}=m(V_8\,\cos 30^\circ+V_c\,\cos 60^\circ) \end{gather*} Equating the above momenta in the $x$-direction and solving for $V_i$, gives \begin{gather*} P_{ix}=P_{fx} \\\\ \Rightarrow V_i=V_8\,\cos 30^\circ+V_c\,\cos 60^\circ \\\\  5=V_8\left(\frac{\sqrt{3}}{2}\right)+V_c\left(\frac{1}{2}\right)\ \ ,\ \ (*) \end{gather*} Next, write the conservation of momentum for the $y$-axis and set $P_{iy}=0$ since the cue ball moves initially only in the $x$ direction. \begin{gather*} P_{iy}=P_{fy}=V_8\,\sin 30^\circ-V_c\,\sin 60^\circ=0 \\\\ V_8\left(\frac{1}{2}\right)-V_c\left(\frac{\sqrt{3}}{2}\right)=0 \\\\ \Rightarrow \quad V_c=\frac{1}{\sqrt{3}}V_8\ \ ,\ \ (**) \end{gather*} Using (*) and (**) and solving for $V_8$, we get: \begin{gather*} 5=V_8\left(\frac{\sqrt{3}}{2}\right)+\frac{1}{\sqrt{3}}V_8\left(\frac{1}{2}\right) \\\\ \Rightarrow V_8=\frac{10}{\sqrt{3}+\frac{1}{\sqrt{3}}}=\boxed{4.33\,\rm m/s}\end{gather*} 
Now by substituting $V_8$ into the relation $*$, we obtain the desired speed $V_C$
\[V_c=\frac{1}{\sqrt{3}}V_8=\frac{10}{4+1}=2.5\,\rm m/s\]

In the collision problems there are two situations:
($i$) Elastic collisions where the linear momentum and kinetic energy are conserved. And we have the following relation $v_{2f}-v_{1f}=v_{1i}-v_{2i}$
($ii$) Inelastic collisions where the linear momentum is conserved but the kinetic energy due to the internal friction is not conserved. In this collision the objects has the same velocity after the collision.
(a) We expect that the lighter ball will rebound to the left after striking heavier ball. Use the conservation of momentum $\vec P_i=\vec P_f$ to find a relation between velocities
\[m_1v_1+m_2v_2=m_1v^{'}_1+m_2v^{'}_2\] 
\[0.25v_1+0=0.25v^{'}_1+2\left(1\right)\Rightarrow v^{'}_1=v_1-8\ ,\ \ \left(1\right)\] 
Using the general relation between relative velocities of the objects in an elastic collision, we get 
\[v^{'}_2-v^{'}_1=v_1-v_2\Rightarrow 1-v^{'}_1=v_1\ ,\ \ (2)\] 
\[\left\{ \begin{array}{rcl}
v^{'}_1 &=& v_1-8\  \\ 
1-v^{'}_1 & = &v_1 \end{array}
\right.\Rightarrow \ v_1-8=1-v_1\Rightarrow 2v_1=9\Rightarrow v_1=+4.5\ \mathrm{m/s}\] 
(b) Because the two balls stick together and have the same velocity after the collision so this is an inelastic collision.
\[m_1v_1+m_2v_2=\left(m_1+m_2\right)v^{'}\] 
\[0.25v_1+0=\left(0.25+2.0\right)\left(1.0\right)\Rightarrow 0.25v_1=2.25\] 
\[v_1=9.0\ \mathrm{m/s}\] 
 

(a) Because after the collision, the two cars stuck together, the collision is inelastic and only the linear momentum of the system is conserved. 
\begin{gather*} \vec{P}_i=\vec{P}_f \\ \Rightarrow m_pv_p+m_c(-v_c)=(m_p+m_c)V \\\\ \to V=\frac{m_p-m_c}{m_p+m_c}v_i \end{gather*} By substituting the numerical values into it, we get \[V=\frac{(3000-1200)}{3000+1200}\times 70=+30\,\rm mph\]
Note: the two cars before the collision have the same velocity but in opposite directions to each other. 

We assume that the pickup moves in the $+x$ and the compact in the $-x$ direction. We see that after the collision the common final velocity of the two cars is positive namely, the cars are moving in the $+x$ direction.

(b) The change in the velocity of pickup is found as below \begin{align*} \Delta v_p&=(v_p)_{after}-(v_p)_{before} \\ &=30-70 \\ &=-40\quad\rm mph \end{align*} And similarly, the change in the compact's velocity is \begin{align*} \Delta v_c&=(v_c)_{after}-(v_c)_{before} \\ &=+30-(-70) \\ &=+100\quad\rm mph\end{align*} 
(c) The changes in the momentum $\Delta P=m\Delta v$ is determined as below \begin{gather*} \Delta P_c=m_c\Delta v_c=1200\,(+100)=12\times {10}^4\,{\rm kg\cdot mph} \\ \Delta P_p=m_p\Delta v_p=3000\,(-40)=-12\times {10}^4\,{\rm kg \cdot mph} \end{gather*} Therefore, the changes in the momentum are the same but in the opposite directions.

(a) Use the conservation of energy just after the bullet exits the block at the speed of $v_2$ and when it reaches the height $h$. That is, $E_i=E_f$
\begin{gather*} E_i=E_f \\\\ \frac{1}{2}mv^2_2=mgh \end{gather*} Solving the above equation for $v_2$ and substituting the given values, we obtain \begin{align*} v_2&=\sqrt{2gh}\\\\&=\sqrt{2\times 9.8\times 1200}\\\\&=153.36\,\rm m/s \end{align*} 
(b) Apply the conservation of the momentum during the collision as \begin{gather*} P_i=P_f \\\\ mv_0=Mv_1+mv_2 \end{gather*} Solving for $v_1$ and plugging in the given numerical values yields: \begin{align*} v_1&=\frac{m}{M}(v_0-v_2)\\\\ &=\frac{0.01}{1}(850-153.36) \\\\ &=6.96\,\rm m/s \end{align*} 
(c) To find the amount of kinetic energy lost during this inelastic collision, subtract the kinetic energies of the bullet and the block from the initial kinetic energy as follows: \begin{align*} \Delta K&=\frac{1}{2}mv^2_0-\left(\frac{1}{2}Mv^2_1+\frac{1}{2}mv^2_2\right) \\\\ &=3470\,\rm J\end{align*} Since $\Delta K\neq 0$, this is an inelastic collision, as expected.
 
(d) When a net average force $\Sigma \vec{F}$ acts on an object during a time interval $\Delta t$, the impulse-momentum theorem states that the impulse of this force equals the change in the momentum of that object, i.e., \[\Sigma \vec{F}\cdot t=\Delta P=m{\vec{v}}_f-m{\vec{v}}_i\] Therefore, substituting the values into this gives: \begin{align*} \Sigma \vec{F}&=\frac{\Delta \vec P}{\Delta t} \\\\ &=\frac{m(v_2-v_0)}{\Delta t} \\\\ &=\frac{(0.01)(153.36-850)}{3\times 10^{-5}} \\\\ &=-2.32\times {10}^5\,\rm N \end{align*} The negative sign indicates that the net force is in the opposite direction of the motion of the bullet (downward). 

Let $v_0$ be the initial velocity, $v_1$ be the velocity of the bullet and block after penetration. 

First, use the conservation of mechanical energy after collision and at the instant that the system reaches the height $8\, \rm{mm}$ to find $v_1$ as
\begin{align*} E_1&=E_2 \\\\ \frac{1}{2}(m+M)v^2_1&=(m+M)gh\\\\ \Rightarrow \quad v_1&=\sqrt{2gh}\\\\&=\sqrt{2\times 9.8\times 0.008}\\\\& =0.396\quad {\rm m/s}\end{align*}
Now use the linear momentum conservation just before and just after the collision. Since the bullet and block have the same velocity immediately after the collision, this case is an inelastic collision. 
\begin{align*} P_0&=P_1\\\\ mv_0&=(m+M)v_1\\\\ \Rightarrow \quad v_0&=\left(1+\frac{M}{m}\right)v_1 \\\\ &=\left(1+\frac{3}{0.024}\right)(0.396) \\\\&=49.9\quad {\rm m/s} \end{align*} Therefore, the initial kinetic energy of the bullet is \begin{align*} K_0&=\frac{1}{2}mv^2_0\\\\&=\frac{1}{2}(0.024)(49.9)^2\\\\&=29.88\quad {\rm J}\end{align*}

To better understand momentum conservation, refer to the solved problems of momentum problems.

The average force acting on an object equals the time rate of change of its linear momentum,i.e., ${\vec{F}}_{ave}=\frac{\Delta \vec{P}}{\Delta t}$ (the original version of Newton's 2nd law), therefore \begin{align*} \Delta \vec{P}&={\vec{P}}_f-{\vec{P}}_i\\\\ &=(0.2)(12(-\hat{i})-20\hat{i}) \\\\ &=-6.4\,\rm kg\cdot m/s \ (\hat{i})\end{align*} Therefore, the average force applied to the ball upon impacting to the wall is calculated as follows: \begin{align*} {\vec{F}}_{ave}&=\frac{6.4}{60\times {10}^{-3}}(-\hat{i}) \\\\ &=106.6\,\rm N\ (-\hat{i})\end{align*} Note: the product of average force and the time interval during which this force is exerted on the body is called the impulse of this force. \[\vec{J}={\vec{F}}_{ave}=\Delta \vec{P}\]
 

First using the conservation of linear momentum find the velocity of the Kirk then use the kinematical relations to determine the desired time.
Since Kirk and jetpack initially are at rest so $P_i=0$, so 
\[{\vec{P}}_i={\vec{P}}_f\Rightarrow 0=m_K{\vec{v}}_K+m_j{\vec{v}}_j\Rightarrow \ {\vec{v}}_K=-\frac{m_j}{m_K}{\vec{v}}_j=-\frac{36.6}{366}\left(-1.55\right)=+0.155\ \frac{\mathrm{m}}{\mathrm{s}}\] 
We have chosen the positive direction to be $+x$ direction.  So Kirk with the constant velocity of $0.155\ \mathrm{m/s\ }$ in direction of the Enterprise moves towards it. Thus, use the definition of the velocity to find the elapsed time.
\[x=x_0+v_0t\Rightarrow t=\frac{124}{0.155}=800s\] 
Convert to minutes as $800\ \mathrm{s\times }\left(\frac{\mathrm{1min}}{\mathrm{60\ s}}\right)=13.3\ \mathrm{min}$
 

First find the energy of the gunpowder that is released and goes into the motion of the ball. 
\[Q=\frac{1}{2}mv^2_0=\frac{1}{2}\left(20\right){\left({10}^3\right)}^2={10}^7\ \mathrm{J}\] 
this energy causes the motion of the ball and cannon (conservation of energy!) so $Q=\frac{1}{2}Mv^2_c+\frac{1}{2}mv^2_b$. Now use the conservation of the momentum to calculate the speed of the cannon after firing then substitute it into the $Q$ relation. 
\[{\vec{P}}_i={\vec{P}}_f\Rightarrow 0=M{\vec{v}}_c+m{\vec{v}}_b\Rightarrow {\vec{v}}_c=-\frac{m}{M}{\vec{v}}_b\] 
\[\Rightarrow \ {\vec{v}}_c=-\frac{20}{500}\left(1000\right)=-40\ \mathrm{m/s}\] 
The negative means that the cannon moves in opposite direction of the ball. Therefore,
\[Q=\frac{1}{2}Mv^2_c+\frac{1}{2}mv^2_b\Rightarrow v_b=\sqrt{\frac{2Q-Mv^2_c}{m}}=\sqrt{\frac{2\times {10}^7-500\times {\left(-40\right)}^2}{20}}=979.79\ \frac{\mathrm{m}}{\mathrm{s}}\] 
 

(a) Because the bullet and block stick together and have the same velocity after the collision, we have an inelastic collision. Use the conservation of momentum to find this velocity \begin{gather*} P_i=P_f \\\\ mv_0=(m+M)v_f \\\\ \Rightarrow v_f=\frac{m}{m+M}v_0 \end{gather*} Substituting the values into the equation above gives \[v_f=\frac{0.010}{0.010+1}(100)=1\,\rm m/s\] 
(b) In an inelastic collision, the kinetic energy is not conserved, but the change in kinetic energy equals the heat generated during this collision, i.e., $\Delta K=-Q$. Therefore, \begin{align*} -Q&=K_f-K_i\\\\ &=\frac{1}{2}(m+M)v^2_f-\frac{1}{2}mv^2_0 \\\\ &=\frac{1}{2}(0.01+1)(1)^2-\frac{1}{2}(0.01)(100)^2 \end{align*} Hence, as a result of collision, the following amount of heat is generated. \[Q=49.495\,\rm J\] The negative sign indicates that the heat is transferred into the surroundings.
 
(c) The heat produced during this collision causes the temperature of the bullet to increase. Thus, using the equation $Q=mc\Delta T$ and solving for $\Delta T$, we have \begin{align*} \Delta T&=\frac{Q}{mC}\\\\ &=\frac{49.495}{(0.010)(128)} \\\\ &=38.66\,\rm ^\circ C\end{align*}
(d) Use the work-energy theorem as $K_2-K_1=W_{net}$. In this case, only the friction force does work on the system so
\begin{gather*} K_2-K_1=-W_f \\\\ 0-\frac{1}{2}(m+M)v^2_1=-\underbrace{{\mu }_k(m+M)g}_{f_k}d \end{gather*} Solving the above expression for $d$, we get \begin{align*} d&=\frac{v^2_1}{2({\mu }_k g)} \\\\ &=\frac{1^2}{2\times 0.2\times 9.8} \\\\ &=25.51\,\rm cm \end{align*}  

(a) The initial and final momentums of the puck are
\[p_{ix}=mv_{ix}\ ,\ \ p_{iy}=0\ \ ,\ \ p_{fy}=mv_{fy}\ ,\ \ p_{fx}=0\] 
One of the consequences of the Newton's 2${}^{nd}$ law is that the time rate of the change of particle's momentum equals the net force acting on it that is $\vec{F}=\frac{\mathrm{\Delta }\vec{P}}{\mathrm{\Delta }t}$. So 
\[F_x=\frac{\mathrm{\Delta }P_x}{\mathrm{\Delta }t}=\frac{P_{fx}-P_{ix}}{\mathrm{\Delta }t}=\frac{0-mv_{ix}}{\mathrm{\Delta }t}=-\frac{8\times 0.35}{0.1}=-28\ \mathrm{N}\] 
\[F_y=\frac{\mathrm{\Delta }P_y}{\mathrm{\Delta }t}=\frac{P_{fy}-P_{iy}}{\mathrm{\Delta }t}=\frac{mv_{fy}-0}{\mathrm{\Delta }t}=\frac{0.35\times 15}{0.1}=52.5\ \mathrm{N}\] 
(b) Recall that if the components of a vector are given, its magnitude and direction w.r.t horizontal axis are determined by 
\[\left|\vec{F}\right|=\sqrt{F^2_x+F^2_y}=\sqrt{{\left(-28\right)}^2+{\left(52.5\right)}^2}=59.5\ \mathrm{N}\] 
\[\theta ={\tan^{-1} \left(\frac{\left|F_y\right|}{\left|F_x\right|}\right)\ }={\tan^{-1} \left(\frac{52.5}{28}\right)\ }=61.9{}^\circ \] 
Since $x$ component is negative and $y$ component is positive so the resultant vector lies in the second quadrant.
 

(a) Use the conservation of the momentum in the $x$ and $y$ directions:
\[P_{ix}=P_{fx}\Rightarrow m_1{\left(v_{1i}\right)}_x+m_2{\left(v_{2i}\right)}_x=m_1{\left(v_{1f}\right)}_x+m_2{\left({\vec{v}}_{2f}\right)}_x\] 
\[\Rightarrow m_1v_{1i}=m_1v_{1f}\,{\cos 45{}^\circ \ }+m_2v_{2f}\,{\cos 30{}^\circ \ }\ \ ,\ \ \mathrm{(I)}\] 
\[P_{iy}=P_{fy}\Rightarrow 0=m_1{\left(v_{1f}\right)}_y+m_2{\left(v_{2f}\right)}_y\] 
\[\Rightarrow -m_1v_{1f}\,{\sin 45{}^\circ \ }+m_2v_{2f}\,{\sin 30{}^\circ \ }=0\ \ ,\ \ \mathrm{(II)}\] 
From $\mathrm{(II)}$ we get
\[v_{2f}=\frac{m_1v_{1f}\,{\sin 45{}^\circ \ }}{m_2\,{\sin 30{}^\circ \ }}\] 
Insert $v_{2f}$into $\mathrm{(I)}$, we obtain 
\[m_1v_{1i}=m_1v_{1f}\,{\cos 45{}^\circ \ }+m_2\left(\frac{m_1v_{1f}\,{\sin 45{}^\circ \ }}{m_2\,{\sin 30{}^\circ \ }}\right){\cos 30{}^\circ \ }\] 
\[v_{1i}=v_{1f}\left({\cos 45{}^\circ \ }+{\sin 45{}^\circ \ }{\cot 30{}^\circ \ }\right)\] 
\[\Rightarrow v_{1f}=\frac{v_{1i}}{\left({\cos 45{}^\circ \ }+{\sin 45{}^\circ \ }{\cot 30{}^\circ \ }\right)}=2.6\ \frac{\mathrm{m}}{\mathrm{s}}\] 
\[v_{2f}=\frac{m_1v_{1f}\,{\sin 45{}^\circ \ }}{m_2\,{\sin 30{}^\circ \ }}=1.8\ \frac{\mathrm{m}}{\mathrm{s}}\] 
(b) To see that whether the collision is elastic or inelastic, we must calculate the initial and final kinetic energy of the pucks before and after the collision.
\[K_i=K_{1i}+K_{2i}=\frac{1}{2}m_1v^2_{1i}+0=0.31\ \mathrm{J}\] 
\[K_f=K_{1f}+K_{2f}=\frac{1}{2}m_1v^2_{1f}+\frac{1}{2}m_2v^2_{2f}=0.17\ \mathrm{J}\] 
Since during this collision some of the kinetic energy lost as a heat into the surrounding so we have an inelastic collision.

(a) By definition, the center of mass of a collection of point particles is 
\[{\vec{r}}_{cm}=\frac{\Sigma m_i{\vec{r}}_i}{M}\] 
Let the mass of ball be $m_b$. Therefore 
\[x_{cm}=\frac{(m_Ax_A+m_Bx_B)+m_br_b}{M+m_b}=\frac{150\left(0.95\right)+6\times 0.2}{150+6}=0.92\ \mathrm{m}\] 
Since the combined mass of the people and platform is given, we must consider it as a new mass $M$ located at the center of the platform $0.95\mathrm{m}$ away from the axis.
(b) Since there are no frictional forces act on the system (external forces) so due to Newton's 2${}^{nd}$ law the center of the mass does not change.
(c) After the ball is caught by the B, all points of the system moving to the left by amount of $x$. So first, calculate the new center of mass and then set it to the previous magnitude! (see figure below)
\[x^{'}_{cm}=\frac{150\left(0.95-x\right)+6\left(1.7-x\right)}{150+6}\] 
\[x^{'}_{cm}=x_{cm}=0.92\Rightarrow x=0.14\ \mathrm{m}\] 

(a) Using the definition of the center of mass of a system of masses $m_i$ at distances $r_i$ from the origin, we get
\[x_{cm}=\frac{\Sigma m_ir_i}{\Sigma m_i}=\frac{\left(m+M\right)x_1+mx_2}{m+M+m}=\frac{\left(60+40\right)\left(0\right)+(40\times 1.5)}{40+60+40}=0.43\ \mathrm{m}\] 
Away from the origin. 
(b) Since there is no external force acting on the two-platform and man system, Newton's second law $\Sigma {\vec{F}}_{ext}=M\frac{d}{dt}{\vec{v}}_{cm}$states that the velocity of the center of mass is constant over time or is conserved. The initial velocity of cm is zero so its final velocity must be also zero. 
\[{\left(v_{cm}\right)}_i={\left(v_{cm}\right)}_f=0\] 
(b) Similar to the reasoning above, since there is no external force, the cm does not move that is ${\left(x_{cm}\right)}_i={\left(x_{cm}\right)}_f=0.43\ \mathrm{m}$
Let us denote the final position of the platforms by $x^{'}_1$ and $x^{'}_2$. At the moment the man lands on the platform two, the position of this platform has not changed yet so 
\[{\left(x_{cm}\right)}_f=\frac{\left(m+M\right)x^{'}_2+mx^{'}_1}{2m+M}\] 
\[=\frac{\left(m+M\right)x^{'}_2+m\left(-\mathrm{\Delta }x\right)}{2m+M}={\left(x_{cm}\right)}_i=0.43\] 
Solving for $\mathrm{\Delta }x$ (shown in the figure), we get
\[\Rightarrow \frac{\left(40+60\right)\left(1.5\right)-40\mathrm{\Delta }x}{2\left(40\right)+60}=0.43\] 
\[\mathrm{\Delta }x=2.245\ \mathrm{cm}\] 
This is the distance of platform one from the origin at the moment the man lands on the second platform. From the figure, we can find the distance between the platforms as
\[d=\mathrm{\Delta }x+1.5=3.745\ \mathrm{m}\] 

(a) Apply the conservation of linear momentum between entering and lodging the bullet in the block ($2$)
\[P_1=P_2\Rightarrow mV_0=\left(M+m\right)V_2\] 
Now apply the conservation of mechanical energy between the instant of entering the bullet into the block ($2$) and after the bullet + block come to a stop ($3$)($K_f=0$)
\[E_2=E_3\Rightarrow \frac{1}{2}\left(m+M\right)V^2_2=\frac{1}{2}kx^2\] 
By combining these two relation, we obtain the initial velocity of the bullet
\begin{align*} \frac{1}{2}(m+M)\left(\frac{m}{m+M}\right)^2V^2 &=\frac{1}{2}kx^2\\\\ \Rightarrow \quad V&=\frac{L}{m}\sqrt{k(m+M)}\\\\ &=\frac{0.03}{0.008}\sqrt{1500(0.008+4)}\\\\&=290.76\quad {\rm m/s}\end{align*} where in above we set $x=L$.
(b) Impulse is defined by the following relation
\[\vec{J}=\underbrace{\vec{F}\cdot\mathrm{\Delta }t}_{impulse}=\mathrm{\Delta }\vec{P}\] 
So we must find the change in momentum of the block + bullet during the time interval stated above. Block+bullet after coming to a stop, start again move in the opposite direction (the elastic potential energy at the end of the path convert to the kinetic energy) Since there is no friction then there is an asymmetry between going and backing paths. Therefore, we must multiply the going result by two to find the total time in which the block+bullet is in contact.
\begin{align*} J_{2\to 3}&=(m+M)(\underbrace{V_3}_{0}-V_2)\\\\&=(m+M)\left(0-\frac{m}{m+M}V\right) \\\\&=-mV\\\\ &=-0.008\times 290.76=2.32\,{\rm N.m}\end{align*} Therefore, the total momentum is \[J_{tot}=J_{2\to 3}+J_{3\to 2}=4.65\quad {\rm N.s}\] 

You can find more solved problems on impulse and momentum, here.