Learn steps to solve conservation of linear momentum in one dimension and two dimension problems. Practice problem when relative velocity after the event is given.

- If net external force on a system is zero, then momentum of the system is conserved.

\(\vec F_{ext}=0\)

\(\dfrac{d\,\vec P_{sys}}{dt}=0\)

i.e., \(\vec P_{sys}\) is constant.

**Steps to solve problems related to conservation of momentum:**

**Step 1: **Identify the system, such that \(\vec F_{ext}=0\) on the system.

**Step 2: **Draw two separate diagrams to show the system before and after the event.

**Step 3: **List the known information and assign variables to the unknown velocities with respect to the ground.

**Step 4: **Write down the momentum of all the bodies in the system, before and after the event.

\(\vec P=m\,\vec v\)

**Note: **Here, \(\vec v\) is the actual velocity of the particle not the relative velocity.

**Step 5: **As \(\vec F_{ext}=0\), initial momentum of the system is equal to final momentum of the system.

\(\vec P_i=\vec P_f\)

**Step 6: **If relative velocity is given, then find the relation between the velocities, for solving the problem.

A \(-\dfrac{20}{3}\hat i\;m/sec\)

B \(-\dfrac{63}{5}\hat i\;m/sec\)

C \(\dfrac{20}{7}\hat i\;m/sec\)

D \(-50\,\hat i\;m/sec\)

- If net external force on a system is zero, then momentum of the system is conserved.

\(\vec F_{ext}=0\)

\(\dfrac{d\,\vec P_{sys}}{dt}=0\)

i.e., \(\vec P_{sys}\) is constant.

**Case 1:** If net external force in \(x-\) direction is zero, then momentum in \(x-\) direction is conserved.

If \(\Sigma\,(F_x)_{ext}=0\)

then \(\dfrac{d}{dt}P_x=0\)

\(\therefore\,P_x\) is constant.

**Case 2:** If net external force in \(y-\) direction is zero, then momentum in \(y-\) direction is conserved.

If \(\Sigma\,(F_y)_{ext}=0\)

then \(\dfrac{d}{dt}P_y=0\)

\(\therefore\,P_y\) is constant.

**Case 3:** If net external force in \(z-\) direction is zero, then momentum in \(z-\) direction is conserved.

If \(\Sigma\,(F_z)_{ext}=0\)

then \(\dfrac{d}{dt}P_z=0\)

\(\therefore\,P_z\) is constant.

**Steps to solve problems related to conservation of momentum:**

**Step 1:** Identify the system, such that \(\vec F_{ext}=0\) on the system.

**Step 2:** Draw two separate diagrams to show the system before and after the event.

**Step 3:** List the known information and assign variables to the unknown velocities with respect to the ground.

**Step 4:** Write down the momentum of all the bodies in the system, before and after the event.

\(\vec P=m\,\vec v\)

**Note:** Here, \(\vec v\) is the actual velocity of the particle not the relative velocity.

**Step 5:** As \(\vec F_{ext}=0\), initial momentum of the system is equal to final momentum of the system.

\(\vec P_i=\vec P_f\)

**Note**

- For simplicity, conservation of momentum can be applied along the \(x,\;y\) and \(z\) axes respectively.
- If we do not know the angle of an unknown velocity, then instead of taking two variable \(v\) and \(\theta\), it is convenient to use \(v_x\) and \(v_y\).

A \(v_x=\dfrac{9}{5}m/s,\;v_y=\dfrac{4}{5}m/s\)

B \(v_x=\dfrac{2}{5}m/s,\;v_y=\dfrac{3}{5}m/s\)

C \(v_x=\dfrac{4}{5}m/s,\;v_y=\dfrac{7}{5}m/s\)

D \(v_x=\dfrac{4}{5}m/s,\;v_y=\dfrac{9}{5}m/s\)

- If net external force on a system is zero, then momentum of the system is conserved.

\(\vec F_{ext}=0\)

\(\dfrac{d\,\vec P_{sys}}{dt}=0\)

i.e., \(\vec P_{sys}\) is constant.

- Thus it means that initial momentum of the system is same as final momentum of the system.

\(\vec P_i=\vec P_f\)

- So, considering the trolley and man as a system.
- Before the man jumps onto a moving trolley, the initial momentum of the system is given as,

\(\vec P_i=m_1u_1+m_2u_2\)

- When the man jumps onto the moving trolley, their relative velocity becomes zero, i.e., the velocity of both man and trolley is same.
- So, final momentum of the system is given as,

\(\vec P_f=(m_1+m_2)v\)

- Since no external force is acting, so momentum of the system is conserved.

\(\vec P_i=\vec P_f\)

**Steps to solve problems related to conservation of momentum:**

**Step 1:** Identify the system, such that \(\vec F_{ext}=0\) on the system.

**Step 2:** Draw two separate diagrams to show the system before and after the event.

**Step 3:** List the known information and assign variables to the unknown velocities with respect to the ground.

**Step 4:** Write down the momentum of all the bodies in the system, before and after the event.

\(\vec P=m\,\vec v\)

**Note:** Here, \(\vec v\) is the actual velocity of the particle not the relative velocity.

**Step 5:** As \(\vec F_{ext}=0\), initial momentum of the system is equal to final momentum of the system.

\(\vec P_i=\vec P_f\)

- If net external forces on a system is zero, then momentum of the system is conserved.

If \(\vec F_{net}=0\)

\(\dfrac{d\vec P_{sys}}{dt}=0\)

i.e., \(\vec P_{sys}\) is constant.

- When one body gets separated from another body with given relative velocity, that is, when the event is instantaneous like firing, jumping, explosion, etc., then relative velocity is defined just after the event.

- The velocity attained by one body is relative to the velocity attained by the second body just after the event as

\(\vec v_{1/2}=\vec v_1-\vec v_2\)

**Steps to solve problem when relative velocity after the event is given:**

- Draw the diagram of two given bodies, before the event and after the event.
- Assign velocity with appropriate directions and unknown velocities with any direction, with respect to the ground.

**Note - **

- Do not try minimize the number of variables.
- Assign variables to unknown velocities, which are further calculated by using equation.

3. For given relative velocity after the event use,

\(\vec v_{1/2}=\vec v_1-\vec v_2\)

**Note - **If it is not mentioned or clarified in the problem, we assume the relative velocity to be after the event.

A \(20\,m/sec\)

B \(25\,m/sec\)

C \(15\,m/sec\)

D \(10\,m/sec\)