According to conservation law of momentum, the impulse is equal to zero.

$$
begin{align*}
Delta p_{t} &= Delta p_{1} + Delta p_{2} \
end{align*}
$$

According to the Momentum-Impulse Theorem:

$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= Delta vec{p} \
&= left( vec{p_{f}} – vec{p_{i}} right) \
end{align*}
$$

So, if the total external force that actingon the system is zero, then the impulse and the changing in momentum is zero, too. Substituting in the previous calculations, then we get

$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= 0 \
&= Delta vec{p} \
&= left( vec{p_{f}} – vec{p_{i}} right) \
vec{p_{f}} &= vec{p_{i}} \
end{align*}
$$

If the total external force that affecting on the system is zero, then the system obeys the conservation law of momentum. So, the initial momentum of the system is equal to the final momentum of the system.

Solve for the elastic collision:

According to conservation law of momentum, the impulse is equal to zero.

$$
begin{align*}
Delta p_{t} &= Delta p_{1} + Delta p_{2} \
end{align*}
$$

According to the Momentum-Impulse Theorem:

$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= Delta vec{p} \
&= left( vec{p_{f}} – vec{p_{i}} right) \
end{align*}
$$

So, if the total external force that acting on the system is zero, then the impulse and the changing in momentum is zero, too. Substituting in the previous calculations, then we get

$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= 0 \
&= Delta vec{p} \
&= left( vec{p_{f}} – vec{p_{i}} right) \
vec{p_{f}} &= vec{p_{i}} \
end{align*}
$$

Multiple both sides by the velocity, then we get

$$
begin{align*}
vec{p_{f}} cdot vec{v_{f}} &= vec{p_{i}} cdot vec{v_{i}} \
end{align*}
$$

In order to evaluate the kinetic energy, we use the following relation:

$$
begin{align*}
K.E &= m ~ v^{2} \
&= m ~ left( vec{v} cdot vec{v} right) \
&= vec{p} cdot vec{v}
end{align*}
$$

Substituting from the previous calculation, then we get

$$
begin{align*}
vec{p_{f}} cdot vec{v_{f}} &= vec{p_{i}} cdot vec{v_{i}} \
K.E_{f} &= K.E_{i} \
end{align*}
$$

If the total external force that affecting on the system is zero, then the system obeys the conservation law of momentum. So, the initial momentum of the system is equal to the final momentum of the system. So, we deduce the conservation law of energy from the conservation law of momentum. Therefore, in the elastic collision, the initial kinetic energy is equal to the final kinetic energy.

Solve for inelastic collision:

If the total external force that affecting the system is zero, then the system obeys the conservation law of momentum. So, the initial momentum of the system is equal to the final momentum of the system. Therefore, in the elastic collision (ideal state), the initial kinetic energy is equal to the final kinetic energy. In the real world, there are no ideal cases. So, in the real world, there is some energy lost in the inelastic collision. But we know that the energy is conservative. Energy is neither destroyed nor created but changes from to another. So, the lost energy in the inelastic collision is transformed into some other forms of energy as thermal, sound, and friction.

According to conservation law of momentum, the impulse is equal to zero.

$$
begin{align*}
Delta p_{t} &= Delta p_{1} + Delta p_{2} \
&= 0 \
Delta p_{1} &= Delta p_{2} \
end{align*}
$$

So, In the ideal collision, the initial momentums are equal to the final momentums. Also, each particle take different path. But in the inelastic collision, the two objects are sticking after the collision.

According to conservation law of momentum, the impulse is equal to zero.

$$
begin{align*}
Delta p_{t} &= Delta p_{1} + Delta p_{2} \
end{align*}
$$

According to the Momentum-Impulse Theorem:

$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= Delta vec{p} \
&= left( vec{p_{f}} – vec{p_{i}} right) \
end{align*}
$$

So, if the total external force that actingon the system is zero, then the impulse and the changing in momentum is zer, too. Substituting in the previous calculations, then we get

$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= 0 \
&= Delta vec{p} \
&= left( vec{p_{f}} – vec{p_{i}} right) \
vec{p_{f}} &= vec{p_{i}} \
end{align*}
$$

The required condition for momentum to be conserved in a collision:

As we mention that the total momentum of the system should be zero and the external forces such as the friction or the gravity or the normal forces should to be zero.

In the real world, there are no ideal cases. So, in the real world, there is some energy lost in the inelastic collision. But we know that the energy is conservative. Energy is neither destroyed nor created but changes from to another. So, the lost energy in the inelastic collision is transformed into some other forms of energy as thermal, sound, and friction. So, if the cars don’t stick together and no energy lost after the collision, then the collision is an elastic collision.

If the total external force that affecting on the system is zero, then the system obeys the conservation law of momentum. So, the initial momentum of the system is equal to the final momentum of the system.

The mass of the first cart is $m_{1} = 0.12 mathrm{~kg}$. The mass of the second cart is $m_{2} = 0.12 mathrm{~kg}$. The initial speed of the first cart is $v_{1,i} = 0.45 mathrm{~m/s}$. The initial speed of the second cart is $v_{2,i} = 0 mathrm{~m/s}$.

$textbf{Required: }$

(a) Finding the initial kinetic energy of the system.

(b) Finding the final kinetic energy of the system.

In order to evaluate the initial kinetic energy of the system, we use the following relation:

$$
begin{align*}
K.E_{t,i} &= K.E_{1,i} + K.E_{2,i} \
&= dfrac{1}{2} ~ m_{1} ~ v_{1,i}^{2} + dfrac{1}{2} ~ m_{2} ~ v_{2,i}^{2} \
&= dfrac{1}{2} times 0.12 mathrm{~kg} times left( 0.45 mathrm{~m/s} right)^{2} + dfrac{1}{2} times 0.12 mathrm{~kg} times left( 0 mathrm{~m/s} right)^{2} \
&= 0.0122 mathrm{~J}
end{align*}
$$

So, the initial kinetic energy of the system is $0.0122 mathrm{~J}$.

Since the two carts stick together after the collision. So, the final speed of the two carts will be the same. According to conservation law of momentum, the impulse is equal to zero.

$$
begin{align*}
Delta p_{t} &= Delta p_{1} + Delta p_{2} \
&= 0 \
&= m_{1} ~ left( vec{v_{f,1}} – vec{v_{i,1}} right) + m_{2} ~ left( vec{v_{f,2}} – vec{v_{i,2}} right) \
&= m_{1} ~ left( vec{v_{f,1}} – vec{v_{i,1}} right) + m_{2} ~ left( vec{v_{f,2}} – 0 right) \
&= left( m_{1} + m_{2} right) ~ vec{v_{f}} + m_{1} ~ vec{v_{i,1}} \
end{align*}
$$

Rearrange and solve for the final speed of the two carts:

$$
begin{align*}
vec{v_{f}} &= – dfrac{ m_{1} ~ vec{v_{i,1}} }{ left( m_{1} + m_{2} right) } \
&= – dfrac{ 0.12 mathrm{~kg} times left( – 0.45 mathrm{~m/s} right) }{ left( 0.12 mathrm{~kg} + 0.12 mathrm{~kg} right) } \
&= 0.225 mathrm{~m/s}
end{align*}
$$

In order to evaluate the final kinetic energy of the system, we use the following relation:

$$
begin{align*}
K.E_{t,f} &= K.E_{1,f} + K.E_{2,f} \
&= dfrac{1}{2} ~ m_{1} ~ v_{1,f}^{2} + dfrac{1}{2} ~ m_{2} ~ v_{2,f}^{2} \
&= dfrac{1}{2} ~ 2 m ~ v_{f}^{2} \
&= m ~ v_{f}^{2} \
&= 0.12 mathrm{~kg} times left( 0.225 mathrm{~m/s} right)^{2} \
&= 6.075 times 10^{-3} mathrm{~J}
end{align*}
$$

So, the final kinetic energy of the system is $6.075 times 10^{-3} mathrm{~J}$.

(b) The final kinetic energy of the system is $6.075 times 10^{-3} mathrm{~J}$.

The mass of the rock is $m_{r} = 0.19 mathrm{~kg}$. The mass of the hammer is $m_{h} = 0.55 mathrm{~kg}$. The initial speed of the hammer is $v_{h,i} = 4.5 mathrm{~m/s}$. The initial speed of the rock is $v_{r,i} = 0 mathrm{~m/s}$.

$textbf{Required: }$

Finding the final speed of the system.

Since the two carts stick together after the collision. So, the final speed of the two carts will be the same. According to the conservation law of momentum, the impulse is equal to zero.

$$
begin{align*}
Delta p_{t} &= Delta p_{h} + Delta p_{r} \
&= 0 \
&= m_{h} ~ left( vec{v_{f,h}} – vec{v_{i,h}} right) + m_{2} ~ left( vec{v_{f,r}} – vec{v_{i,r}} right) \
&= m_{h} ~ left( vec{v_{f,h}} – vec{v_{i,h}} right) + m_{2} ~ left( vec{v_{f,r}} – 0 right) \
&= left( m_{h} + m_{r} right) ~ vec{v_{f}} + m_{h} ~ vec{v_{i,h}} \
end{align*}
$$

Rearrange and solve for the final speed of the two carts:

$$
begin{align*}
vec{v_{f}} &= dfrac{ m_{h} ~ vec{v_{i,h}} }{ left( m_{h} + m_{r} right) } \
&= dfrac{ 0.55 mathrm{~kg} times 4.5 mathrm{~m/s} }{ left( 0.55 mathrm{~kg} + 0.19 mathrm{~kg} right) } \
&= 3.345 mathrm{~m/s}
end{align*}
$$

So, the final speed of the whole system is $3.345 mathrm{~m/s}$.

The mass of the red car is $m_{1} = 950 mathrm{~kg}$. The initial speed of the red car is $v_{1,i} = 20 mathrm{~m/s}$ in the positive $X$ direction. The mass of the blue minvan is $1300 mathrm{~kg}$. The direction in which the wrecked vehicles move after the collision relative to the positive $X$ direction is $40^{circ}$.

$textbf{Required: }$

Finding the initial speed of the blue minivan.

As the textbook mentions that the momentum is given by the mass times the velocity. So, the total momentum in the $X$ direction is given by

$$
begin{align*}
p_{X} &= left( m_{1} ~ v_{f} + m_{2} ~ v_{f} right) ~ cos left( theta right) \
&= m_{1} ~ v_{1,i} + m_{2} ~ v_{2,i} \
&= m_{1} ~ v_{1,i} + 0 \
&= m_{1} ~ v_{1,i}
end{align*}
$$

Also, the total momentum in the $Y$ direction is given by

$$
begin{align*}
p_{Y} &= left( m_{1} ~ v_{f} + m_{2} ~ v_{f} right) ~ sin left( theta right) \
&= m_{1} ~ v_{1,i} + m_{2} ~ v_{2,i} \
&= 0 + m_{2} ~ v_{2,i} \
&= m_{2} ~ v_{2,i}
end{align*}
$$

To get the direction in which the wrecked vehicles move after the collision relative to the positive $X$ direction, we should divide both equations together.

$$
begin{align*}
dfrac{ m_{2} ~ v_{2,i} }{ m_{1} ~ v_{1,i} } &= dfrac{ left( m_{1} ~ v_{f} + m_{2} ~ v_{f} right) ~ sin left( theta right) }{ left( m_{1} ~ v_{f} + m_{2} ~ v_{f} right) ~ cos left( theta right) } \
&= dfrac{ sin left( theta right) }{ cos left( theta right) } \
&= tan left( theta right)
end{align*}
$$

Rearrange and solve for the initial speed of the blue minivan:

$$
begin{align*}
v_{2,i} &= dfrac{ m_{1} ~ v_{1,i} ~ tan left( theta right) }{ m_{2} } \
&= dfrac{ 950 mathrm{~kg} times 20 mathrm{~m/s} times tan left( 40^{circ} right) }{ 1300 mathrm{~kg} } \
&= 12.264 mathrm{~m/s}
end{align*}
$$

So, the initial speed of the blue minivan is $12.264 mathrm{~m/s}$.

The mass of the Brittany and the skateboard is $m_{1} = 61 mathrm{~kg}$. The initial speed of Brittany is $v_{1,i} = 0 mathrm{~m/s}$. The mass of the pumpkin is $m_{2} = 3.7 mathrm{~kg}$.The final speed of the system (the skateboard) is $v_{f} = 0.16 mathrm{~m/s}$.

$textbf{Required: }$

Finding the initial speed of the pumpkin.

Since the two carts stick together after the collision. So, the final speed of the two carts will be the same. According to the conservation law of momentum, the impulse is equal to zero.

$$
begin{align*}
Delta p_{t} &= Delta p_{1} + Delta p_{2} \
&= 0 \
&= m_{1} ~ left( vec{v_{f,1}} – vec{v_{i,1}} right) + m_{2} ~ left( vec{v_{f,2}} – vec{v_{i,2}} right) \
&= m_{1} ~ left( vec{v_{f,1}} – vec{v_{i,1}} right) + m_{2} ~ left( vec{v_{f,2}} – 0 right) \
&= left( m_{1} + m_{2} right) ~ vec{v_{f}} + m_{2} ~ vec{v_{i,2}} \
end{align*}
$$

Rearrange and solve for the initial speed of the pumpkin:

$$
begin{align*}
vec{v_{i,2}} &= dfrac{ left( m_{1} + m_{2} right) ~ vec{v_{f}} }{m_{2} } \
&= dfrac{ left( 61 mathrm{~kg} + 3.7 mathrm{~kg} right) times 0.16 mathrm{~m/s} }{ 3.7 mathrm{~kg} } \
&= 2.798 mathrm{~m/s}
end{align*}
$$

So, the initial speed of the pumpkin is $2.798 mathrm{~m/s}$.