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Collision

Practice elastic collision in two dimensions and equation of inelastic collision. Learn elastic and inelastic collision formula and definition by practice problems.

Elastic and Inelastic Collision

Collision

  • The event in which two bodies tend to occupy same position at the same time is called collision.
  • A collision is an event in which two bodies exert force on each other for a very short duration.
  • When two bodies collide with each other , the shape of the bodies deform for a very short time interval.
  • This effect can be seen through high speed cameras.
  • After the impact ,some bodies regain their original shape and some do not.

Elastic Collision

  • The bodies which regain their original shape after collision are called elastic bodies and collision is known as elastic collision.

Inelastic Collision

  • The bodies which do not regain their original shape after collision are called inelastic bodies and collision is known as inelastic collision. 

Illustration Questions

Which among the following represents elastic collision?

A

B

C

D

×

After collision, if the shape of the body remains same, then collision is elastic collision.

The shape of the body before and after collision is same in only option  (A).

Hence, only option (A) is correct which shows elastic collision.

Which among the following represents elastic collision?

A image
B image
C image
D image

Option A is Correct

Mathematical Representation of Elastic Collision 

  • On the basis of kinetic energy we can differentiate between elastic and inelastic collision.
  • If the kinetic energy before collision is same as kinetic energy after collision, then it is called elastic collision.
  • If a spring is fully unscratched after collision, the whole potential energy will be converted into kinetic energy of the system.
  • Since, no external force is involved during collision, the momentum of the system remains constant.

Momentum Before Collision  = Momentum During Collision  = Momentum After Collision 

  • As elastic collision occurs between elastic bodies, the loss in kinetic energy during collision converts into work done by conservative forces i.e., potential energy of system.
  • Thus the total energy will be converted into its kinetic energy after collision.

Mathematical Representation of Elastic Collision

  • Consider two bodies of mass m1 and m2 , moving with velocity v1 and v2 respectively.
  • Since, the body regains its shape after collision, it is elastic collision.
  • Kinetic energy before collision

\(\Rightarrow KE_{BC} = \dfrac{1}{2} m_1\,u_1^2 + \dfrac{1}{2} m_2\,u_2^2\)

  • Kinetic energy after collision

\(\Rightarrow KE_{AC} = \dfrac{1}{2} m_1\,v_1^2 + \dfrac{1}{2} m_2\,v_2^2\)

  • So, kinetic energy before collision is same as kinetic energy after collision.

\(\Rightarrow K.E_{BC} = K.E_{AC}\neq K.E_{DC}\)         

 \(\Rightarrow \dfrac{1}{2}m_1\,u^2_1+\dfrac{1}{2}m_2\,u^2_2=\dfrac{1}{2}m_1\,v^2_1+\dfrac{1}{2}m_2\,v^2_2\)            ..........(1)

Momentum Conservation

  • Since, no external force is applied during collision so, momentum of the system remains conserved.

Momentum Before Collision  = Momentum During Collision  = Momentum After Collision 

     \(\Rightarrow m_1\,u_1+m_2\,u_2=m_1\,v+m_2\,v=m_1\,v_1+m_2\,v_2\)   ........(2)

Illustration Questions

Among the following which equation is incorrect for elastic collision?

A \(m_1\,u_1 + m_2\,u_2 = m_1\,v+m_2\,v\)

B \(m_1\,v_1 + m_2\,v_2 = m_1\,v+m_2\,v\)

C \(\dfrac{1}{2}m_1\,u_1^2 + m_2\,u_2^2 = \dfrac{1}{2} m_1\,v^2+m_2\,v^2\)

D \(\dfrac{1}{2}m_1\,u_1^2 + \dfrac{1}{2}m_2\,u_2^2 = \dfrac{1}{2} m_1\,v_1^2+\dfrac{1}{2}m_2\,v_2^2\)

×

Since, no external force is applied during collision so, momentum of the system remains conserved.

Momentum Before Collision  = Momentum During Collision  = Momentum After Collision 

Also, by using sign convention

      \(m_1u_1 + m_2u_2 = m_1\,v + m_2\,v = m_1 v_1 +m_2 \,v_2 \)

Hence, option (A) and (B) are correct. 

Kinetic energy before collision is same as kinetic energy after collision.

 \(\Rightarrow K.E_{BC} = K.E_{AC} \neq K.E_{DC}\)  

\(\Rightarrow \dfrac{1}{2} m_1\,u_1^2 +\dfrac{1}{2} m_2\,u_2^2 = \dfrac{1}{2} m_1\,v_1^2 +\dfrac{1}{2} m_2\,v_2^2\)

Hence, option (C) is incorrect.

Kinetic energy before collision is same as kinetic energy after collision.

\(\Rightarrow K.E_{BC} = K.E_{AC} \neq K.E_{DC}\)  

\(\Rightarrow \dfrac{1}{2} m_1\,u_1^2 +\dfrac{1}{2} m_2\,u_2^2 = \dfrac{1}{2} m_1\,v_1^2 +\dfrac{1}{2} m_2\,v_2^2\)  

Hence, option (D) is correct.

Among the following which equation is incorrect for elastic collision?

image
A

\(m_1\,u_1 + m_2\,u_2 = m_1\,v+m_2\,v\)

.

B

\(m_1\,v_1 + m_2\,v_2 = m_1\,v+m_2\,v\)

C

\(\dfrac{1}{2}m_1\,u_1^2 + m_2\,u_2^2 = \dfrac{1}{2} m_1\,v^2+m_2\,v^2\)

D

\(\dfrac{1}{2}m_1\,u_1^2 + \dfrac{1}{2}m_2\,u_2^2 = \dfrac{1}{2} m_1\,v_1^2+\dfrac{1}{2}m_2\,v_2^2\)

Option C is Correct

Inelastic Collision

  • In inelastic collision, there are non-conservative forces within the body.
  • Due to non- conservative force, the whole kinetic energy used to change the shape cannot be regained as kinetic energy lost in sound and heat etc.
  • Thus, the final kinetic energy of the system will be less than initial kinetic energy.
  • But as there is no external force, the momentum of the system remains constant.
  • So, mathematically 

           \(K.E_{\text {After collision}} <K.E_ {\text {Before collision}} \)

and  \(p_{\text {Before collision}} =p_{\text {During collision}} =p_{\text {After collision}}\) 

Where   \(K.E={\text {Kinetic energy}}\) 

\(\Rightarrow p={\text {momentum}}\)  

Illustration Questions

Choose correct relation for inelastic collision.

A \(m_1\,u_1 - m_2\,v_2 = m_1\,v +m_2\,v\)

B \(\dfrac{1}{2}\,m_1\,u_1^2 +\dfrac{1}{2} m_2\,u_2^2 = \dfrac{1}{2}m_1\,v^2 +\dfrac{1}{2} m_2\,v^2\)

C \(\dfrac{1}{2}\,m_1\,u_1^2 +\dfrac{1}{2} m_2\,u_2^2 < \dfrac{1}{2}m_1\,v^2_1 +\dfrac{1}{2} m_2\,v^2_2\)

D \(\dfrac{1}{2}\,m_1\,u_1^2 +\dfrac{1}{2} m_2\,u_2^2 > \dfrac{1}{2}m_1\,v^2_1 +\dfrac{1}{2} m_2\,v^2_2\)

×

Since, no external force is acting on the body.

Hence, momentum of the system remain conserved. 

Also, by using sign convention,

\(\Rightarrow m_1\,u_1-m_2\,u_2=-m_1\,v-m_2\,v=- m_1\,v_1-m_2\,v_2\)

Hence, option (A) is incorrect.

In inelastic collision, the final kinetic energy of the system is less then the initial kinetic energy.

\(\Rightarrow\dfrac{1}{2} m_1\,u_1^2+\dfrac{1}{2} m_2\,u_2^2>\dfrac{1}{2} m_1\,v_1^2+\dfrac{1}{2} m_2\,v_2^2\)

Hence, (B) and (C) are incorrect and option (D) is correct.

Choose correct relation for inelastic collision.

image
A

\(m_1\,u_1 - m_2\,v_2 = m_1\,v +m_2\,v\)

.

B

\(\dfrac{1}{2}\,m_1\,u_1^2 +\dfrac{1}{2} m_2\,u_2^2 = \dfrac{1}{2}m_1\,v^2 +\dfrac{1}{2} m_2\,v^2\)

C

\(\dfrac{1}{2}\,m_1\,u_1^2 +\dfrac{1}{2} m_2\,u_2^2 < \dfrac{1}{2}m_1\,v^2_1 +\dfrac{1}{2} m_2\,v^2_2\)

D

\(\dfrac{1}{2}\,m_1\,u_1^2 +\dfrac{1}{2} m_2\,u_2^2 > \dfrac{1}{2}m_1\,v^2_1 +\dfrac{1}{2} m_2\,v^2_2\)

Option D is Correct

Condition for Elastic Collision 

  • On the basis of kinetic energy we can differentiate between elastic and inelastic collision.
  • If the kinetic energy before collision is same as kinetic energy after collision, then it is called elastic collision.
  • As elastic collision occurs between elastic bodies, the loss in kinetic energy during collision converts into work done by conservative forces i.e., potential energy of system.
  • Thus the total energy will be converted into its kinetic energy after collision.
  • If a spring is fully unscratched after collision, the whole potential energy will be converted into kinetic energy of the system.
  • Since, no external force is involved during collision, the momentum of the system remains constant.

Momentum Before Collision  = Momentum During Collision  = Momentum After Collision 

Mathematical Representation of Elastic Collision

  • Consider two bodies of mass m1 and m2, moving with velocity v1 and v2 respectively.
  • Since, the body regains its shape after collision, it is elastic collision.
  • Kinetic energy before collision

?\(\Rightarrow KE_{BC} = \dfrac{1}{2} m_1\,u_1^2 + \dfrac{1}{2} m_2\,u_2^2\)

  • Kinetic energy after collision

\(\Rightarrow KE_{AC} = \dfrac{1}{2} m_1\,v_1^2 + \dfrac{1}{2} m_2\,v_2^2\)      

  • So, kinetic energy before collision is same as kinetic energy after collision.

\(\Rightarrow K.E_{BC} = K.E_{AC}\neq K.E_{DC}\)         

\(\Rightarrow \dfrac{1}{2}m_1\,u^2_1+\dfrac{1}{2}m_2\,u^2_2=\dfrac{1}{2}m_1\,v^2_1+\dfrac{1}{2}m_2\,v^2_2\)          ..........(1)

Momentum Collision 

  • Since, no external force is applied during collision so, momentum of the system remains conserved.

           Momentum Before Collision  = Momentum During Collision  = Momentum After Collision 

     \(\Rightarrow m_1\,u_1+m_2\,u_2=m_1\,v+m_2\,v=m_1\,v_1+m_2\,v_2\)   ........(2)

  • Thus from equation (1) and (2) we get,

\(\Rightarrow m_1\,u_1 +m_2\,u_2 = m_1\,v_1 +m_2\,v_2 \)              .......(3)

and \(\dfrac{1}{2}m_1\,u_1^2 +m_2\,u_2^2 =\dfrac{1}{2} m_1\,v_1^2 +\dfrac{1}{2}m_2\,v_2^2 \)  ......(4)

Calculation of vand v2

  • Rewriting these equations, we get 

          \(\Rightarrow m_1(u_1-v_1) = m_2(u_2-v_2) \)  .....(5)

and   \(m_1(u_1^2-v_1^2) = m_2(u_2^2-v_2^2) \) 

\(\Rightarrow m_1(u_1-v_1)(u_1 +v_1) = m_2(v_2-u_2)(v_2 +u_2)\) .........(6)

  • Dividing equation (6) by (5), we get

\(\Rightarrow u_1 +v_1 = v_2+u_2\)

or, \( u_1 -u_2 = v_2-v_1\)

or,  \(( u_1 -u_2) = -(v_1-v_2)\) 

Conclusion 

Thus, in elastic collision the relative velocity of particles before collision is the negative of the relations velocity after collision.

  • Also by solving equation (3) and (4) we get,

      \(\Rightarrow v_1 = \left(\dfrac{m_1-m_2}{m_1+m_2}\right)\,u_1+ \left(\dfrac{2\,m_2}{m_1+m_2}\right)\,u_2\)      .......(7)

and   \( v_2 = \left(\dfrac{2\,m_1}{m_1+m_2}\right)\,u_1+ \left(\dfrac{\,m_2-m_1}{m_1+m_2}\right)\,u_2\)     ........(8)

  • Let us consider that  \(m_1\) is much grater than \(m_2\), i.e., \(m_1>>>m_2\) 

         then, from equation (7)

                 \(v_1 = u_1\) 

and from equation (8), 

\(\Rightarrow v_2 = 2\,u_1 - u_2\)

Conclusion

The velocity of a heavier body remains unchanged during collision.

Illustration Questions

Calculate the value of velocity \(v_2\) if the given collision is elastic collision.

A \(5 \,m/sec\)

B \(6\,m/sec\)

C \(11 \,m/sec\)

D \(2 \,m/sec\)

×

In elastic collision the relative velocity of particles before collision is the negative of the relative velocity after collision.

\((u_1-u_2) = - (v_1-v_2)\)

where,  \(u_1=\)  velocity of A before collision

 \(u_2=\)  velocity of B before collision

 \(v_1=\)  velocity of A after collision

 \(v_2=\)  velocity of B after collision

Given : \(u_1 = 10\,m/sec, \,\,\,u_2 = 5\,m/sec , \,\,\,v_1 =6\,m/sec\)  

So, \((10-5) = -(6-v_2)\)

\(\Rightarrow 5=\, – 6 + v_2 \)

\(\Rightarrow v_2 = 5+6 = 11\, m/sec\)

Calculate the value of velocity \(v_2\) if the given collision is elastic collision.

image
A

\(5 \,m/sec\)

.

B

\(6\,m/sec\)

C

\(11 \,m/sec\)

D

\(2 \,m/sec\)

Option C is Correct

Elastic Collision 

  • On the basis of kinetic energy we can differentiate between elastic and inelastic collision.
  • If the kinetic energy before collision is same as kinetic energy after collision, then it is called elastic collision.
  • As elastic collision occurs between elastic bodies , the loss in kinetic energy during collision converts into work done by conservative forces i.e., potential energy of system.
  • Thus the total energy will be converted into its kinetic energy after collision.
  • If a spring is fully unscratched after collision, the whole potential energy will be converted into kinetic energy of the system.
  • Since, no external force is involved during collision, the momentum of the system remains constant.

Momentum Before Collision  = Momentum During Collision  = Momentum After Collision 

Mathematical Representation of Elastic Collision

  • Consider two bodies of mass m1 asd m2, moving with velocity v1 and v2 respectively.
  • Since, the body regains its shape after collision, it is elastic collision.
  • Kinetic energy before collision

\(\Rightarrow KE_{BC} = \dfrac{1}{2} m_1\,u_1^2 + \dfrac{1}{2} m_2\,u_2^2\)

  • Kinetic energy after collision

\(\Rightarrow KE_{AC} = \dfrac{1}{2} m_1\,v_1^2 + \dfrac{1}{2} m_2\,v_2^2\)

  • So, kinetic energy before collision is same as kinetic energy after collision.

             \(\Rightarrow K.E_{BC} = K.E_{AC}\neq K.E_{DC}\)

             \(\Rightarrow \dfrac{1}{2}m_1\,u^2_1+\dfrac{1}{2}m_2\,u^2_2=\dfrac{1}{2}m_1\,v^2_1+\dfrac{1}{2}m_2\,v^2_2\)    ..........(1)

Momentum Collision

Since, no external force is applied during collision so, momentum of the system remains conserved.

Momentum Before Collision  = Momentum During Collision  = Momentum After Collision 

 \(\Rightarrow m_1\,u_1+m_2\,u_2=m_1\,v+m_2\,v=m_1\,v_1+m_2\,v_2\)   ........(2)

  • Thus from equation (1) and (2) we get,

           \(\Rightarrow m_1\,u_1 +m_2\,u_2 = m_1\,v_1 +m_2\,v_2 \)   .......(3)

and \(\dfrac{1}{2}m_1\,u_1^2 +m_2\,u_2^2 =\dfrac{1}{2} m_1\,v_1^2 +\dfrac{1}{2}m_2\,v_2^2 \)    ......(4)

Calculation of  \(v_1 \,\,\text{and}\,\,v_2\) 

  • Rewriting these equations, we get 

          \(\Rightarrow m_1(u_1-v_1) = m_2(u_2-v_2) \)   .....(5)

and   \(m_1(u_1^2-v_1^2) = m_2(u_2^2-v_2^2) \) 

\(\Rightarrow m_1(u_1-v_1)(u_1 +v_1) = m_2(v_2-u_2)(v_2 +u_2)\) ........(6)

  • Dividing equation (6) by  (5), we get

\(\Rightarrow u_1 +v_1 = v_2+u_2\)

or, \( u_1 -u_2 = v_1-v_2\)

or,  \(( u_1 -u_2) = -(v_1-v_2)\) 

Conclusion 

Thus, in elastic collision the  relative velocity of particles before collision is the negative of the relations velocity after collision.

  • Also  by solving equation (3) and (4) we get,

      \(\Rightarrow v_1 = \left(\dfrac{m_1-m_2}{m_1+m_2}\right)\,u_1+ \left(\dfrac{2\,m_2}{m_1+m_2}\right)\,u_2\)      .......(7)

and   \( v_2 = \left(\dfrac{2\,m_1}{m_1+m_2}\right)\,u_1+ \left(\dfrac{\,m_2-m_1}{m_1+m_2}\right)\,u_2\)     ........(8)

  • Let us consider that  \(m_1\) is much grater than \(m_2\), i.e., \(m_1>>>m_2\) 

         then, from equation (7)

  \(v_1 = u_1\) 

and from equation (8), 

\(\Rightarrow v_2 = 2\,u_1 - u_2\)

  • Let the masses of both bodies are same, i.e,

\(m_1 = m_2 =m\)           

The from equitation (7) and (8)

\(\Rightarrow v_1 = u_2\)

and  \( v_2 = u_1\)

Conclusion

If the masses of two bodies are same, then the velocity of both bodies will get inter changed after collision.   

Illustration Questions

Choose the incorrect option among the following regarding elastic collision.

A

B

C

D

×

If the masses of two bodies are same, then the velocity of both bodies will get inter changed after collision.  

Hence, option (B) is incorrect.

Choose the incorrect option among the following regarding elastic collision.

A image
B image
C image
D image

Option B is Correct

Two Dimensional Collision

  • When the line of velocity of two bodies after collision is different, it is known as 2- Dimensional collision.

Momentum in 2- Dimensional Elastic Collision

  • In elastic collision, momentum in x- axis and y- axis are individually conserved as no external force is acting along x- axis or y- axis.

Kinetic Energy in 2- Dimensional Elastic Collision 

  • In elastic collision, the kinetic energy before and after collision remains same.
  • Consider an example as shown in figure.
  • Resolving velocities along x- axis and y- axis, we get

\(\Rightarrow \dfrac{1}{2} m_1\,u_1^2 + 0 = \dfrac{1}{2} m_1\,v_1^2+\dfrac{1}{2} m_2\,v_2^2\)  .........(1)  [kinetic energy is conserved]

\(\Rightarrow m\,u_1 = m_1\,v_1 \,cos\,\theta_1+m_2\,v_2 \,cos\,\theta_2\)     .......(2)     [along x- axis]

\(\Rightarrow 0+0 = m_1\,v_1 \,sin\,\theta_1-m_2\,v_2 \,sin\,\theta_2\)      .......(3)   [along y- axis]

Illustration Questions

Choose the incorrect relation among the following elastic collision.

A \(\dfrac{1}{2} m_1\,u^2 = \dfrac{1}{2} m_1\,v_1^2+\dfrac{1}{2} m_2\,v_2^2\)

B \(m\,v_1=m_1\,v_1 \,cos\,\theta_1+m_2\,v_2 \,cos\,\theta_2\)

C \(0=m_1\,v_1 \,sin\,\theta_1-m_2\,v_2 \,sin\,\theta_2\)

D \(\dfrac{1}{2} m_1\,v^2 = \dfrac{1}{2} m_1\,v_1^2\,cos^2\,\theta_1+\dfrac{1}{2} m_2\,v_2^2\,cos^2\,\theta_2\)

×

In elastic collision, the kinetic energy before and after collision remains same.

So,  \(\dfrac{1}{2} m_1\,u^2 =\dfrac{1}{2} m_1\,v_1^2+\dfrac{1}{2} m_2\,v_2^2 \) 

Hence, option (A) is correct and option (D) is incorrect. 

In elastic collision, momentum in x- axis and y- axis are individually conserved as no external force is acting along x- axis or y- axis.

Hence, along x- axis,

\(\Rightarrow m\,v_1 = m_1\,v_1\,cos\,\theta_1 +m_2\,v_2 \,cos\,\theta_2\)

and along y- axis,

\(\Rightarrow 0 = m_1\,v_1 \,sin\,\theta_1 - m_2\,v_2 \,sin\,\theta_2\)

Hence, option (B) and (C) are correct.

Choose the incorrect relation among the following elastic collision.

image
A

\(\dfrac{1}{2} m_1\,u^2 = \dfrac{1}{2} m_1\,v_1^2+\dfrac{1}{2} m_2\,v_2^2\)

.

B

\(m\,v_1=m_1\,v_1 \,cos\,\theta_1+m_2\,v_2 \,cos\,\theta_2\)

C

\(0=m_1\,v_1 \,sin\,\theta_1-m_2\,v_2 \,sin\,\theta_2\)

D

\(\dfrac{1}{2} m_1\,v^2 = \dfrac{1}{2} m_1\,v_1^2\,cos^2\,\theta_1+\dfrac{1}{2} m_2\,v_2^2\,cos^2\,\theta_2\)

Option D is Correct

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