Informative line

Conservation Of Linear Momentum

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.

Conservation of Linear Momentum in 1 Dimension

Condition for conservation of momentum

  • 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.

Illustration Questions

A bomb of mass, \(m=10\,kg\) at rest suddenly explodes into two pieces of masses \(m_1=4\,kg\) and \(m_2=6\,kg\) respectively. If the velocity \((v_1)\) of \(m_1\) is \(10\hat i\,m/s\,\), then find the velocity of \(m_2\).

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

×

Let us consider the system that includes \(m_1\) and \(m_2\).

image

Assigning velocity with respect to ground frame \(u=0\;m/sec,\;\;\;v_1=10\;m/sec\)

image

Initial momentum of the system,

\((\vec P_i)_m=m×\vec u\)

\(=10×0\)

\(=0\,kg.m/s\)

image

Final momentum of the system

\((\vec P_f)_m=(\vec P_f)_{m_1}+(\vec P_f)_{m_2}\)

\(=m_1×\vec v_1+m_2×\vec v_2\)

\(=4×10\,\hat i+6×v_2\,\hat i\)

\(=(40\,\hat i+6\,v_2\hat i)\,kg.m/s\)

image

Since there is no external force acting on the body, so momentum of the system is conserved.

So, \(P_i=P_f\)

\(0=40\,\hat i+6\,v_2\,\hat i\)

\(\vec v_2=\dfrac{-40}{6}\,\hat i\)

\(\vec v_2=\dfrac{-20}{3}\,\hat i\,m/sec\)

image

A bomb of mass, \(m=10\,kg\) at rest suddenly explodes into two pieces of masses \(m_1=4\,kg\) and \(m_2=6\,kg\) respectively. If the velocity \((v_1)\) of \(m_1\) is \(10\hat i\,m/s\,\), then find the velocity of \(m_2\).

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

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

Option A is Correct

Conservation of Linear Momentum in 2 Dimension

Condition for conservation of momentum

  • 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 \(v_y\).

Illustration Questions

A bomb of mass \(m=10\;kg\), kept at rest, suddenly explodes into three pieces of masses \(m_1=2\,kg,\;m_2=3\,kg\) and \(m_3=5\,kg\). If \(m_1\) and \(m_2\) move in perpendicular directions with speed \(2\,m/s\) and \(3\,m/s\) respectively, then find the velocity of \(5\,kg\) mass.  

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

×

Given system consists of a bomb and the three parts of the bomb formed after explosion.

image

Assign velocity with respect to ground frame,

\(u=0\,m/s,\;v_1=2\,m/s,\;v_2=3\,m/s\)

\(v_x=x-\)components of velocity of mass \(m_3\)

\(v_y=y-\)components of velocity of mass \(m_3\)

When two perpendicular directions are to be taken, it is convenient to assume them along \(x\) and \(y-\)axes.

image

Initial momentum of the system

\((\vec P_i)_m=m×\vec u\)

\(=10×0\)

\(=0\;kg\,m/s\)

image

Final momentum of the system

\((\vec P_f)_m=(\vec P_f)_{m_1}+(\vec P_f)_{m_2}+(\vec P_f)_{m_3}\)

\(=(m_1\vec v_1)+(m_2\vec v_2)+(m_3\vec v_3)\)

\(=(2×2\,\hat j)+(3×3\,\hat i)+5(-v_x\,\hat i-v_y\,\hat j)\)

\(=4\,\hat j+9\,\hat i-5v_x\hat i-5v_y\hat j\)

image

Since, there is no external force acting on the body so momentum of the system is conserved.

\(\vec P_i=\vec P_f\)

\(\Rightarrow\;0=(9-5v_x)\,\hat i+(4-5v_y)\,\hat j\)

\(\therefore\;9-5v_x=0\)

\(v_x=\dfrac{9}{5}m/s\)

and,

\(4-5v_y=0\)

\(v_y=\dfrac{4}{5}m/s\)

image

A bomb of mass \(m=10\;kg\), kept at rest, suddenly explodes into three pieces of masses \(m_1=2\,kg,\;m_2=3\,kg\) and \(m_3=5\,kg\). If \(m_1\) and \(m_2\) move in perpendicular directions with speed \(2\,m/s\) and \(3\,m/s\) respectively, then find the velocity of \(5\,kg\) mass.  

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

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

Option A is Correct

Conservation of Momentum

Condition for conservation of momentum

  • 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.

\(\vec P_i=\vec P_f\)

 

Illustration Questions

A trolley of mass \(m_2=80\,kg\) is moving with a velocity \(u_2=5\,m/s.\) A man of mass \(m_1=40\,kg\), running at a speed \(u_1=10\,m/s\), chasing the trolley, jumps onto it. Find the speed of the trolley after the jump. Neglect friction.

A \(10\,m/s\)

B \(6.66\,m/s\)

C \(5\,m/s\)

D \(0\,m/s\)

×

Let us consider a system consisting of a man and trolley.

image

Assigning velocities with respect to ground frame

\(u_1=10\,m/s,\;\;u_2=5\,m/s\)

image

Initial momentum of the system

\((\vec P_i)_1=m_1u_1\)

\(=40×10\)

\(=400\,kg\;m/s\)

\((\vec P_i)_2=m_2u_2\)

\(=80×5\)

\(=400\,kg\;m/s\)

image

Final momentum of the system

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

\(=(40+80)v\)

\(=120v\;kg\,m/s\)

image

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

\(\vec P_i=\vec P_f\)

\((400+400)=120v\)

\(v=\dfrac{800}{120}\)

\(v=\dfrac{20}{3}m/s\)

\(v=6.66\,m/s\)

image

A trolley of mass \(m_2=80\,kg\) is moving with a velocity \(u_2=5\,m/s.\) A man of mass \(m_1=40\,kg\), running at a speed \(u_1=10\,m/s\), chasing the trolley, jumps onto it. Find the speed of the trolley after the jump. Neglect friction.

A

\(10\,m/s\)

.

B

\(6.66\,m/s\)

C

\(5\,m/s\)

D

\(0\,m/s\)

Option B is Correct

Relative Velocity After the Event

Condition for conservation of momentum

  • 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.

Relative velocity in an event

  • 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.

Relative velocity 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:

  1. Draw the diagram of two given bodies, before the event and after the event.
  2. 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.

Illustration Questions

Rita, having mass m is sitting on a trolley of mass M, moving with, \(u_0 = 10\,m/s\). Suddenly she jumps off the trolley with a velocity, \(v_{rel}= 20\,m/s\) with respect to the final velocity of trolley. If the velocity of trolley, after the jump is \(v_0 = 5\,m/sec\) then, find velocity of Rita.

A \(20\,m/sec\)

B \(25\,m/sec\)

C \(15\,m/sec\)

D \(10\,m/sec\)

×

Consider Rita and trolley as a system. 

image

Given: \(u_0 = 10\,m/sec, \,\,\,v_0 = 5 \,m/s\)

All velocities are given with respect to ground frame

image

The relative velocity is given with respect to the final velocity of the trolley, i.e., after the event.

\(\therefore \vec v_{rel} = \vec v_1 - \vec v_0\)

where \(v_1\) is the velocity of Rita after jump

\(v_0\) is the velocity of trolley after jump

image

Given, \(\vec v_{rel} = 20 \,m/sec\) 

\(\vec v_0 = 5\,m/sec\)

\(\therefore \vec v_1 = \vec v_{rel}+ \vec v_0\)

\( = 20+5\)

\( = 25 \,m/sec\)

image

Rita, having mass m is sitting on a trolley of mass M, moving with, \(u_0 = 10\,m/s\). Suddenly she jumps off the trolley with a velocity, \(v_{rel}= 20\,m/s\) with respect to the final velocity of trolley. If the velocity of trolley, after the jump is \(v_0 = 5\,m/sec\) then, find velocity of Rita.

A

\(20\,m/sec\)

.

B

\(25\,m/sec\)

C

\(15\,m/sec\)

D

\(10\,m/sec\)

Option B is Correct

Practice Now