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

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

So, the total momentum is twice the initial momentum $p_{t} = p_{i} + p_{f} = 2 p_{i}$.

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

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

So, the internal forces acting on each object individually. So, each object has a momentum in a random direction which makes the total momentum of the system is equal to zero. Because the moments are equal in magnitude and in the opposite directions.

Solve for the two skaters at rest:

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

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

Since the two skaters push off each other in opposite directions. So, the total force is equal to zero. Also, the total momentum of the system is equal to zero. The friction force should to be neglected to consider the system is ideal.

Solve for the surface is frictionless surface:

Since the friction force is neglected, Then no energy will lose within the collision.

According to the Momentum-Impulse Theorem:

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

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

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

So, the initial and the final momentum are equals. Therefore, the answer is “No, it is impossible for both of the objects at rest after collision.”

Solve for the surface is frictionless surface:

Since the friction force is neglected, Then no energy will lose within the collision.

According to the Momentum-Impulse Theorem:

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

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

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

So, the initial and the final momentum are equals. Therefore, the answer is “Yes, it is possible for one of the objects at rest after collision.”

According to the Momentum-Impulse Theorem:

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

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

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

So, when you drop your keys, their momentum increases (their momentum is not conserved). So, there is an external force that acts on the keys to make them fall. Then Earth exerts an opposite gravitational force that cancels the reacting force on the keys (due to Newton’s third law). So, there is momentum generating in the opposite direction of the keys’ momentum that leads to momentum conservation. So, the momentum of the universe is conservative. Therefore, the answer is “No, the momentum of the universe doesn’t increase as the keys fall.”

The mass of the first canoe is $m_{1} = 320 mathrm{~kg}$. The initial speed of the two canoes is equal to zero. The final speed of the first canoe is $v_{f,1} = 0.58 mathrm{~m/s}$. The final speed of the second canoe is $v_{f,2} = 0.42 mathrm{~m/s}$.

$textbf{Required: }$

Finding the mass of the second canoe.

According to the Momentum-Impulse Theorem:

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

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

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

Rearrange and solve for the mass of the second canoe:

$$
begin{align*}
m_{2} &= – dfrac{ m_{1} ~ v_{f,1} }{ v_{f,2} } \
&= – dfrac{ 320 mathrm{~kg} times 0.58 mathrm{~m/s} }{ – 0.42 mathrm{~m/s} } \
&= 441.905 mathrm{~kg}
end{align*}
$$

So, the mass of the second canoe is $441.905 mathrm{~kg}$.

The mass of the lumberjack is $m_{1} = 94 mathrm{~kg}$. The mass of the water’s pool is $m_{2} = 1.2 times 10^{3} mathrm{~kg}$. The initial speed is equal to zero. The final speed of the lumberjack is $v_{1} = 2.2 mathrm{~m/s}$.

$textbf{Required: }$

Finding the final speed of the log relative to the shore.

According to the Momentum-Impulse Theorem:

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

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

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

Rearrange and solve for the final speed of the log relative to the shore:

$$
begin{align*}
v_{f,2} &= – dfrac{ m_{1} ~ v_{f,1} }{ m_{2} } \
&= – dfrac{ 94 mathrm{~kg} times left( – 2.2 mathrm{~m/s} right) }{1.2 times 10^{3} mathrm{~kg} } \
&= 0.172 mathrm{~m/s}
end{align*}
$$

So, the final speed of the log relative to the shore is $0.172 mathrm{~m/s}$.

The mass of the ice holding helmet is $m_{1} = 1.3 mathrm{~kg}$. The initial speed of the both the helmet and the player is equal to zero. The final speed of the helmet is $v_{f,1} = 6.5 mathrm{~m/s}$. The final speed of the helmet after it recoils is $v_{f,2} = 0.25 mathrm{~m/s}$.

$textbf{Required: }$

Finding the mass of the hockey player.

According to the Momentum-Impulse Theorem:

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

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

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

Rearrange and solve for the mass of the hokey player:

$$
begin{align*}
m_{2} &= – dfrac{ m_{1} ~ v_{f,1} }{ v_{f,2} } \
&= – dfrac{ 1.3 mathrm{~kg} times 6.5 mathrm{~m/s} }{ – 0.25 mathrm{~m/s} } \
&= 33.8 mathrm{~kg}
end{align*}
$$

So, the mass of the hockey player is $33.8 mathrm{~kg}$.