The first exerted force on the car is $F_{1} = 305 mathrm{~N}$. The time interval for the first impulse is $t_{1} = 6 mathrm{~s}$. The second exerted force on the car is $F_{2} = 215 mathrm{~N}$. The time interval for the second impulse is $t_{2} = 6 mathrm{~s}$.

$textbf{Required: }$

(a) Finding the final momentum of the car.

(b) Finding the final speed of the car.

According to the Momentum-Impulse Theorem:

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

In order to evaluate the total impulse on the car, we use the following relation:

$$
begin{align*}
I_{t} &= I_{1} + I_{2} \
&= vec{F_{1}} ~ Delta t_{1} + vec{F_{2}} ~ Delta t_{2} \
&= 305 mathrm{~N} times 6 mathrm{~s} + 215 mathrm{~N} times 6 mathrm{~s} \
&= 3.12 times 10^{3} mathrm{~kg cdot m/s}
end{align*}
$$

Rearrange and solve for the final momentum of the object:

$$
begin{align*}
vec{p_{f}} &= vec{I} + vec{p_{i}} \
&= 3.12 times 10^{3} mathrm{~kg cdot m/s} + 0 mathrm{~kg cdot m/s} \
&= 3.12 times 10^{3} mathrm{~kg cdot m/s} \
end{align*}
$$

So, the final momentum of the object is $3.12 times 10^{3} mathrm{~kg cdot m/s}$.

Solve for the final speed:

In order to evaluate the magnitude of the Momentum of the baseball, we use the formula that is given in the second yellow box:

$$
begin{align*}
p &= m ~ v \
end{align*}
$$

From example $7.6$, the mass of the car and the driver is $1360 mathrm{~kg}$.

Rearrange and solve for the final speed:

$$
begin{align*}
v_{f} &= dfrac{ p_{f} }{ m }\
&= dfrac{ 3.12 times 10^{3} mathrm{~kg cdot m/s} }{1360 mathrm{~kg} } \
&= 2.294 mathrm{~m/s}
end{align*}
$$

So, the final speed is $2.294 mathrm{~m/s}$.

(b) The final speed of the car is $2.294 mathrm{~m/s}$.

The mass of the baseball is $m = 0.144 mathrm{~kg}$. The speed of the baseball is $v = 43 mathrm{~m/s}$. The average exerted force by the bat on the baseball is $F = 6.5 times 10^{3} mathrm{~N}$. The time interval is $Delta t = 1.22 times 10^{-3} mathrm{~s}$.

$textbf{Required: }$

Finding the final speed of the ball.

As the textbook assigns that the impulse is the product of a force and the time which is given by

$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= 6.5 times 10^{3} mathrm{~N} times 1.22 times 10^{-3} mathrm{~s} \
&= 7.93 mathrm{~kg cdot m/s}
end{align*}
$$

According to the Momentum-Impulse Theorem:

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

Rearrange and solve for the final momentum of the object:

$$
begin{align*}
vec{p_{f}} &= vec{I} + vec{p_{i}} \
&= vec{I} + m ~ vec{v} \
&= 7.93 mathrm{~kg cdot m/s} – 0.144 mathrm{~kg} times 43 mathrm{~m/s} \
&= 1.738 mathrm{~kg cdot m/s} \
end{align*}
$$

In order to evaluate the final momentum, we use the following relation:

$$
begin{align*}
vec{p_{f}} &= m ~ vec{v_{f}} \
end{align*}
$$

Rearrange and solve for the final speed:

$$
begin{align*}
v_{f} &= dfrac{ p_{f} }{m} \
&= dfrac{ 1.738 mathrm{~kg cdot m/s} }{ 0.144 mathrm{~kg} } \
&= 12.069 mathrm{~m/s}
end{align*}
$$

So, the final speed of the baseball is $12.069 mathrm{~m/s}$.

The initial velocity of the ball is $v_{i} = 4.2 mathrm{~m/s}$. The final velocity of the ball is $v_{f} = – 24 mathrm{~m/s}$. The impulse delivered to the ball by the player is $I = – 9.3 mathrm{~kg cdot m/s}$.

$textbf{Required: }$

Finding the ball’s mass.

As the textbook assigns that the impulse is the product of a force and the time which is given by

$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
end{align*}
$$

According to the Momentum-Impulse Theorem:

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

Rearrange and solve for the mass of the ball $m$:

$$
begin{align*}
m &= dfrac{ vec{I} }{ left( vec{v_{f}} – vec{v_{i}} right)} \
&= dfrac{ – 9.3 mathrm{~kg cdot m/s} }{ left( – 24 mathrm{~m/s} – 4.2 mathrm{~m/s} right)} \
&= 0.3298 mathrm{~kg}
end{align*}
$$

So, the mass of the ball is $0.3298 mathrm{~kg}$.

The changing of the momentum of the falling orange per second is $Delta p = 2.4 mathrm{~kg cdot m/s}$.

$textbf{Required: }$

(a) Finding the weight of the free-falling orange.

(b) Finding the time interval that the orange takes to fall which the changing momentum of the orange is $Delta p = 6.1 mathrm{~kg cdot m/s}$.

Since the orange is free-falling, Therefore, the only force that acts on the orange is the weight of the orange.

As the textbook assigns that the impulse is the product of a force and the time which is given by

$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= W ~ Delta t \
end{align*}
$$

Rearrange and solve for the weight of the orange:

$$
begin{align*}
W &= dfrac{ vec{I} }{ Delta t } \
&= dfrac{ Delta vec{p} }{ Delta t } \
&= dfrac{ 2.4 mathrm{~kg cdot m/s} }{ 1 mathrm{~s} } \
&= 2.4 mathrm{~N}
end{align*}
$$

So, the weight of the orange is $2.4 mathrm{~N}$ downward.

Since the orange is free-falling, Therefore, the only force that acts on the orange is the weight of the orange.

As the textbook assigns that the impulse is the product of a force and the time which is given by

$$
begin{align*}
vec{I} &= vec{F} ~ Delta t \
&= W ~ Delta t \
end{align*}
$$

Rearrange and solve for the time interval:

$$
begin{align*}
Delta t &= dfrac{ vec{I} }{W } \
&= dfrac{ Delta vec{p} }{ W } \
&= dfrac{ 6.1 mathrm{~kg cdot m/s} }{ 2.4 mathrm{~N} } \
&= 2.542 mathrm{~s}
end{align*}
$$

So, the time interval that the orange takes to fall is $2.542 mathrm{~s}$.

(b) The time interval that the orange takes to fall is $2.542 mathrm{~s}$.