In this article, we will learn about collisions in detail. We will start with what collisions are and then move on to their types and then to 5 solved numerical problems.

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What are Collisions?

In Physics, Collisions are the strong interactions among bodies involving an exchange of momentum in a short time interval.

Types of Collision

Based on the conservation of kinetic energy, collisions are classified into

(i) Elastic Collision

(ii) Inelastic Collision

Elastic Collision

It is the type of collision in which both the momentum and the kinetic energy are conserved.

{\left( {{{\bf{P}}_{sy}}} \right)_i} = {\left( {{{\bf{P}}_{sy}}} \right)_f}\,;\quad {\left( {K{E_{sy}}} \right)_i} = {\left( {K{E_{sy}}} \right)_f}

Forces involved during collisions are conservative in nature.

Examples:

  • A collision between atomic particles
  • A collision between smooth billiard balls
  • Collision of \alpha  - particle with nucleus.

Inelastic Collision

It is the collision in which the energy is not conserved. The momentum, however, is conserved.

Ex: Collision between two vehicles.

Perfectly inelastic Collisions

It is the collision in which the colliding bodies stick together and move as a single body after the collision. In a perfectly inelastic collision, the momentum remains conserved, but the loss of kinetic energy is maximum.

Example: A bullet fired into the wooden block and remained embedded.

 

Collisions: Important terms

Line of impact

The line passing through the common normal to the surface in contact during impact is called the line of impact. The force during collision act along this on both bodies.

Coefficient of restitution

Newton introduced a dimensionless parameter called the coefficient of restitution (e) to measure the elasticity of collision. It is defined as the ratio of the relative velocity of separation to the relative velocity of approach of the two colliding bodies.

e = \frac{{{\rm{Relative-velocity-of-separation}}}}{{{\rm{Relative-velocity-of-approach}}}}

\Rightarrow \quad \left( {{{\bf{v}}_2} - {{\bf{v}}_1}} \right) = e\left( {{{\bf{u}}_1} - {{\bf{u}}_2}} \right)

This formula is applied along the line of impact. Here, the velocities mentioned in the expression should be taken along the line of impact.

For perfectly elastic collision, e = 1.

For an inelastic collision, 0 < e < 1.

For perfectly inelastic collision e = 0.

 

 

Head-on elastic collision

Let two bodies of masses {m_1} and {m_2} moving with initial velocities {{\bf{u}}_1} and {{\bf{u}}_2} ({{\bf{u}}_1} > {{\bf{u}}_2}). After collision two bodies will move with velocity {{\bf{v}}_1} and {{\bf{v}}_2}.

From the law of conservation of linear momentum,

{m_1}{{\bf{u}}_1} + {m_2}{{\bf{u}}_2} = {m_1}{{\bf{v}}_1} + {m_2}{{\bf{v}}_2}     …(1)

And by the definition of coefficient of restitution,

({{\bf{v}}_2} - {{\bf{v}}_1}) = e({{\bf{u}}_1} - {{\bf{u}}_2})                     …(2)

Solving (1) & (2), for {{\bf{v}}_1} and {{\bf{v}}_2},

{{\bf{v}}_1} = \left( {\frac{{{m_1} - e{m_2}}}{{{m_1} + {m_2}}}} \right){{\bf{u}}_1} + \left( {\frac{{(1 + e){m_2}}}{{{m_1} + {m_2}}}} \right){{\bf{u}}_2}

{{\bf{v}}_2} = \left( {\frac{{(1 + e){m_1}}}{{{m_1} + {m_2}}}} \right){{\bf{u}}_1} + \left( {\frac{{{m_2} - e{m_1}}}{{{m_1} + {m_2}}}} \right){{\bf{u}}_2}

Loss in kinetic energy of system :

\Delta KE = {\left( {K{E_{sy}}} \right)_i} - {\left( {K{E_{sy}}} \right)_f}

\Delta KE = \frac{1}{2}\left( {\frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}} \right){\left| {{{\bf{u}}_1} - {{\bf{u}}_2}} \right|^2}(1 - {e^2})

In the case of perfectly inelastic collision, e = 0

Loss in KE of system is \Delta KE = \frac{1}{2}\left( {\frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}} \right){\left( {\left| {{{\bf{u}}_1} - {{\bf{u}}_2}} \right|} \right)^2}

If two bodies are approaching each other, then the loss in KE of the system is maximum.

{\left[ {{{(\Delta KE)}_{sys}}} \right]_{\max }} = \frac{1}{2}\left( {\frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}} \right){({u_1} + {u_2})^2}

 

Oblique collisions (or collisions in two dimensions)

(1) A pair of equal and opposite impulses act along a common normal direction. Hence linear momentum of individual particles changes along the common normal direction.

(2) No component of impulse acts along the common tangent direction. Hence linear momentum (or) linear velocity of individual particles remains unchanged along this direction.

(3) Net impulse on the particles is zero during the collision. Hence, the net momentum of both the particles remains conserved before and after collision in any direction.

(4) The equation for the coefficient of restitution can be applied along the common normal direction or the line of impact.

From law of conservation of linear momentum along x-axis,

{m_1}{u_1}\cos {\theta _1} + {m_2}{u_2}\cos {\theta _2} = {m_1}{v_1}\cos {\beta _1} + {m_2}{v_2}\cos {\beta _2}

Along y-axis,

{m_1}{u_1}\sin {\theta _1} + {m_2}{u_2}\sin {\theta _2} = {m_1}{v_1}\sin {\beta _1} + {m_2}{v_2}\sin {\beta _2}

Using the velocity of separation = e (velocity of approach) along the line of impact.

\left( {{v_2}\cos {\beta _2} - {v_1}\cos {\beta _1}} \right) = e\left( {{u_1}\cos {\theta _1} - {u_2}\cos {\theta _2}} \right)

 

 

Solved problems on Collisions

 

Problem 1: Elastic head-on collision

Two bodies of masses {m_1} and {m_2} are moving with velocities of 1 m/s and 3 m/s, respectively, in opposite directions. If the bodies undergo one-dimensional elastic collision, the body of mass {m_1} comes to rest. Find the ratio of {m_1} and {m_2}.

Answer

Let bodies {m_1} and {m_2} as a system

Since, ({{\bf{F}}_{ext}}) = 0\,;\,\,{({{\bf{P}}_{sys}})_i} = {({{\bf{P}}_{sys}})_f}

\Rightarrow \quad {m_1}{{\bf{v}}_1} + {m_2}{{\bf{v}}_2} = {m_1}{\bf{v}}_1^{\bf{'}} + {m_2}{\bf{v}}_2^{\bf{'}}

\Rightarrow \quad {m_1}(1\,{\rm{m/s}})\hat i + {m_2}(3\,{\rm{m/s}})( - \hat i) = {m_1} \times 0 + {m_2}{\bf{v}}_2^{\bf{'}}

\Rightarrow \quad {m_1}\hat i - 3{m_2}\hat i = {m_1}{\bf{v}}_2^{\bf{'}}                 …(1)

Using, {\bf{v}}_2^{\bf{'}} - {\bf{v}}_1^{\bf{'}} = e\,({{\bf{v}}_1} - {{\bf{v}}_2})

\Rightarrow \quad {\bf{v}}_2^{\bf{'}} - 0 = 1\,\left( {1\hat i - 3( - \hat i)} \right)

\Rightarrow \quad {\bf{v}}_2^{\bf{'}} = 4\,\hat i                                               …(2)

From (1) & (2),

{m_1}\hat i - 3{m_2}\hat i = 4{m_2}\hat i\quad  \Rightarrow \quad {m_1} = 7{m_2}

\Rightarrow \quad \frac{{{m_1}}}{{{m_2}}} = \frac{7}{1}.

 

 

Problem 2: Inelastic head-on collision

Ball 1 collides with another identical ball 2 at rest as shown in the figure. For what value of the coefficient of restitution e, the velocity of the second ball becomes two times that of the first ball after the collision.

Answer

Since both balls are identical therefore, {m_1} = {m_2}.

Let initial speeds of balls 1 and 2 are {v_1} and {v_2} and after collision their speeds are v_1^{\bf{'}} and v_2^{\bf{'}}.

Using momentum conservation ; \sum\limits_{}^{} {{{\bf{F}}_{ext}}}  = 0.

{m_1}{{\bf{v}}_1} + {m_2}{{\bf{v}}_2} = {m_1}{\bf{v}}_1^{\bf{'}} + {m_2}{\bf{v}}_2^{\bf{'}}

Since, {m_1} = {m_2}\,;\,\,{v_2} = 0\,;\,\,{\bf{v}}_2^{\bf{'}} = 2{\bf{v}}_1^{\bf{'}}

{{\bf{v}}_1} + 0 = {\bf{v}}_1^{\bf{'}} + {m_2}{\bf{v}}_2^{\bf{'}}                        …(1)

Using, {\bf{v}}_2^{\bf{'}} - {\bf{v}}_1^{\bf{'}} = e\left( {{{\bf{v}}_1} - {{\bf{v}}_2}} \right)

\Rightarrow \quad 2{\bf{v}}_1^{\bf{'}} - {\bf{v}}_1^{\bf{'}} = e\left( {{{\bf{v}}_1} - 0} \right)

\Rightarrow \quad {\bf{v}}_1^{\bf{'}} = e\,{{\bf{v}}_1}

Now, from (1),

{\bf{v}}_1^{\bf{'}} = e\,3{\bf{v}}_1^{\bf{'}}\quad  \Rightarrow \quad e = \frac{1}{3}.

 

 

Problem 3: Inelastic Oblique collision

After a perfectly inelastic collision between two identical particles moving at the same speed but in different directions, the speed of the combined particle becomes half the initial speed of either particle. The angle between the velocities of the two before the collision is …

Answer

System: both particles ;

Here \left( {{{\bf{F}}_{ext}}} \right) acting on the system = 0

{{\bf{P}}_1} + {{\bf{P}}_2} = {\bf{P}}

m{{\bf{v}}_1} + m{{\bf{v}}_2} = 2m{{\bf{v}}_f}                  …(1)

Here, \left| {{{\bf{v}}_1}} \right| = \left| {{{\bf{v}}_2}} \right|; let {v_1} = {v_2} = {v_0} then

{v_f} = \frac{{{v_0}}}{2}

From (1),

{(m{v_0})^2} + {(m{v_0})^2} + 2m{v_0}m{v_0}\cos \theta  = {\left( {2m\frac{{{v_0}}}{2}} \right)^2}

\Rightarrow \quad {m^2}v_0^2 + {m^2}v_0^2 + 2{m^2}v_0^2\cos \theta  = {(2m)^2}\frac{{v_0^2}}{4}

\Rightarrow \quad v_0^2 + v_0^2 + 2v_0^2\cos \theta  = v_0^2

\Rightarrow \quad 2v_0^2\cos \theta  =  - v_0^2

\Rightarrow \quad \cos \theta  =  - \frac{1}{2}\quad  \Rightarrow \quad \theta  = {120^o}.

 

 

Problem 4: Energy loss in a head-on collision

A block of mass 0.5 kg is moving with a speed of 2.0 m/s on a smooth surface. It strikes a stationary mass of 1.00 kg head-on, and then they move together as a single body. Find the energy loss during the collision.

Answer

System: both mass ;