91Ó°ÊÓ

Given that a railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car initially has a total mass (car and contents) of 200 kg and is travelling east with a velocity of magnitude 5.00 m/s.

Let mass thrown out be$m_A=25$ kg

And mass of car is$m_B=175$ kg

Initial velocity of car$u=5$ m/s East

According to law of conservation of momentum, the momentum of a system is always conserved if no external force is applied to the system.

Let$v_1$ be the velocity of car after 25 kg mass is thrown out

Velocity of throwing out the mass sideways with respect to car is$v_A=5$ m/s

According to conservation of momentum

role="math" localid="1667975389128" $$\begin{gathered} \left(m_A+m_B\right)u=m_A{v}_A+m_B{v}_1 \\ \left(200\right)\left(5\right)=\left(25\right)\left(5\right)+175v_1 \\ {v}_1=\frac{1000-125}{175} \\ {v}_1=5 \end{gathered}$$

Hence,the final velocity of the car is 5 m/s, East.

Let$v_2$ be the velocity of car after 25 kg mass is thrown out

Velocity of throwing 25 kg out of the car backways relative to car is $v_A=-5$m/s

Velocity of car is 5 m/s

So velocity of mass relative to ground is -5+5 = 0 m/s

According to the conservation of momentum

$$\begin{gathered} \left(m_A+m_B\right)u=m_A{v}_A+m_B{v}_2 \\ \left(200\right)\left(5\right)=\left(25\right)\left(5\right)+175v_2 \\ {v}_2=\frac{1000}{175} \\ {v}_2=5.71 \end{gathered}$$

Hence, the final velocity of the car is 5.71 m/s, East.

Let$v_3$ be the velocity of the car after m = 25 kg mass is thrown out

Velocity of throwing the mass into the carrelative to the ground and opposite in direction to the initial velocity of the car is role="math" localid="1667975644761" $u_A=-6$m/s

According to conservation of momentum

$$\begin{gathered} m{u}_A+\left(m_A+m_B\right)u_B=\left(m_A+m_B\right)v_3 \\ \left(25\right)\left(-6\right)=\left(200\right)\left(5\right)+225v_3 \\ {v}_3=3.78 \end{gathered}$$

Hence, the final velocity of the car is 3.78 m/s, East.