Collision of Bodies MCQ Quiz - Objective Question with Answer for Collision of Bodies - Download Free PDF

Concept:

When two bodies collide and move together after the collision, the principle of conservation of momentum is applied.

The equation for conservation of momentum is given by:

\( m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \)

Where:

• \( m_1 = 3 \) kg (mass of first sphere)
• \( v_1 = 5 \) m/s (velocity of first sphere before collision)
• \( m_2 = 2 \) kg (mass of second sphere)
• \( v_2 = 0 \) m/s (velocity of second sphere before collision, since it is at rest)
• \( v_f \) is the common velocity after collision

Calculation:

Applying the conservation of momentum equation:

\( (3 \times 5) + (2 \times 0) = (3 + 2) v_f \)

\( 15 = 5 v_f \)

\( v_f = \frac{15}{5} = 3 \) m/s

Concept:

Two bodies are connected by a string over a pulley. One mass \( m_1 \) is on a rough horizontal surface, and another mass \( m_2 \) hangs freely.

Let coefficient of friction be \( \mu \). We apply Newton's second law to both bodies.

For mass \( m_1 \) on surface:

Tension - Friction = \( m_1 a \Rightarrow T - \mu m_1 g = m_1 a \)

For hanging mass \( m_2 \):

Weight - Tension = \( m_2 a \Rightarrow m_2 g - T = m_2 a \)

Adding both equations:

\( m_2 g - \mu m_1 g = m_1 a + m_2 a \)

\( \Rightarrow a = \frac{g(m_2 - \mu m_1)}{m_1 + m_2} \)

This is the acceleration of the system.

Concept:

In an elastic collision, momentum is conserved. When two objects collide elastically, their total momentum before and after the collision remain the same.

• The momentum of the system before collision equals the momentum of the system after collision.

Given:

Mass of first ball, m₁ = 0.5 kg

Initial velocity of first ball, u₁ = 10 m/s

Mass of second ball, m₂ = 1 kg

Initial velocity of second ball, u₂ = 0 m/s

Final velocity of first ball, v₁ = 0 m/s

To Find:

Final velocity of the second ball, v₂

Using conservation of momentum:

\({m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2}\)

Substituting the known values:

\({(0.5 \, \text{kg} \times 10 \, \text{m/s}) + (1 \, \text{kg} \times 0 \, \text{m/s}) = (0.5 \, \text{kg} \times 0 \, \text{m/s}) + (1 \, \text{kg} \times v_2)}\)

\({5 \, \text{kg} \cdot \text{m/s} = 1 \, \text{kg} \times v_2}\)

⇒ v₂ = 5 m/s

∴ The final velocity of the second ball after the collision is 5 m/s.

Explanation:

Elastic collision:

• In an elastic collision, both linear momentum and kinetic energy are conserved.
• In other words, the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision, and the total linear momentum of the system is conserved.
• For a two-body elastic collision, the conservation of linear momentum can be expressed as:

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

Where, m₁ and m₂ are the masses of the two bodies, v_1i and v_2i are their initial velocities, v_1f and v_2f are their final velocities.

Inelastic collision:

• During an inelastic collision, two or more bodies collide and stick together, resulting in a single mass with a common velocity.
• The loss of kinetic energy is often associated with the internal forces and deformations that occur during the collision.
• In an inelastic collision, kinetic energy is not conserved.
• Unlike elastic collisions, where the total kinetic energy of the system remains constant, in inelastic collisions, some of the initial kinetic energy is transformed into other forms of energy, such as internal energy or deformation.
• For a two-body inelastic collision, the conservation of linear momentum is still valid, just as in elastic collisions.
• The conservation of linear momentum in a two-body inelastic collision can be expressed as:

m1v1i + m2v2i = (m1 + m2)v_f

Where, m1 and m2 are the masses of the two bodies, v1i and v2i are their initial velocities, v_f is the final velocity after collision.

• The total kinetic energy always decreases in an inelastic collision, the individual kinetic energy of one object might increase at the expense of others, due to the internal forces and deformations during the collision.
• However, even in such cases, the overall kinetic energy of the specific body will always be lower after the collision compared to its pre-collision value.

Concept:

The energy dissipation during impact is called by the term, coefficient of restitution, a scalar quantity

\(e = \frac{{velocity\;of\;separation}}{{Velocity\;of\;approach}}\)

\(e = \frac{{{v_1} - {v_2}}}{{{u_2} - {u_1}}}\)

where, v = velocity of the body after impact, u = velocity of the body before impact

• For perfectly elastic collision, e = 1
• For inelastic collision, e < 1
• For a perfectly inelastic collision, e = 0

Concept:

After the impact the height attained by the ball (h₂) can be calculated by:

h₂ = e² × h₁

where e is coefficient of restitution and h₁ is the initial height from which the ball is dropped

Calculation:

Given:

h₁ = 10 m, h₂ = 2.5 m

We know that

h2 = e2 × h1

2.5 m = e² × 10

e² = 0.25 

e = 0.5 

Explanation:

Types of collision:

Perfectly elastic collision: A perfectly elastic collision is defined as one in which there is no loss of kinetic energy and the momentum of the system is conserved in the collision.

Inelastic collision: A inelastic collision is defined as one in which there is a loss of kinetic energy and the momentum of the system is conserved in the collision.

The coefficient of restitution is the ratio of relative velocity after impact to the relative velocity before impact. 

Coefficient of restitution (e)

\({\rm{e}} = \frac{{{\rm{Relative\;velocity\;after\;collision}}}}{{{\rm{Relative\;velocity\;before\;collision}}}} = \frac{{{{\rm{v}}_2} - {{\rm{v}}_1}}}{{{{\rm{u}}_1} - {{\rm{u}}_2}}}\)

• For perfectly elastic collision, e = 1
• For inelastic collision, e < 1
• For a perfectly inelastic collision, e = 0

Concept:

The energy dissipation during impact is called by the term, coefficient of restitution, a scalar quantity

\(e = \frac{{velocity\;of\;separation}}{{Velocity\;of\;approach}}\)

\(e = \frac{{{v_1} - {v_2}}}{{{u_2} - {u_1}}}\)

where, v = velocity of the body after impact, u = velocity of the body before impact

• For perfectly elastic collision, e = 1
• For inelastic collision, e < 1
• For a perfectly inelastic collision, e = 0

Note:

When two bodies collide they exchange energy and obviously the body with the higher energy will transfer some of its energy to the one with low energy. Hence the body which was having higher velocity initially will now have the lower velocity. 

Concept:

Perfectly elastic collision:

If the law of conservation of momentum and that of kinetic energy hold good during the collision.

Inelastic collision:

If the law of conservation of momentum holds good during a collision while that of kinetic energy is not.

Coefficient of restitution (e)

\({\rm{e}} = \frac{{{\rm{Relative\;velocity\;after\;collision}}}}{{{\rm{Relative\;velocity\;before\;collision}}}} = \frac{{{{\rm{v}}_2} - {{\rm{v}}_1}}}{{{{\rm{u}}_1} - {{\rm{u}}_2}}}\)

• For perfectly elastic collision, e = 1
• For inelastic collision, e < 1
• For perfectly inelastic collision, e = 0

Concept:

This is the case of plastic impact as both balls stick after the collision. 

Plastic impact/ Perfectly inelastic impact

• The two bodies move together with common velocity after the impact
• Momentum is conserved, i.e. m1u1 + m2u2 = (m1 + m2) v0, where v₀ is common velocity after impact.
• There is a loss in kinetic energy (K.E) after the impact.

Let, the two perfectly elastic spherical bodies are A and B of mass m_A, & m_B the initial velocity of ball A  is uA, the initial velocity of ball B is uB, v0 is common velocity after impact.

Moment of conservation:

m_AuA + m_BuB = (m_A + m_B) v0

Calculation:

Given:

m_A = 4 kg, m_B = 6 kg, u_A = 5 m/s, v₀ = 8 m/s

mAuA + mBuB = (mA + mB) v0

(4 × 5) + (6 × u_B) = (6 + 4) × 8

u_B = 10 m/s

∴ the velocity of ball B is 10 m/s.

Explanation:

Direct Impact:

• If the motion of two colliding bodies is directed along the line of impact, the impact is said to be the direct impact.

• In Head-on Inelastic Collision only linear momentum remains constant. v₁ , v₂ are initial velocities and \(v_1^, , v_2^,\) are final velocities of the colliding bodies.

The loss in Kinetic energy during In-elastic collision is given by:

\({\rm{\Delta }}E = \frac{{{m_1}{m_2}}}{{2\left( {{m_1} + {m_2}} \right)}}{\left( {{v_1} - {v_2}} \right)^2}(1-e^2)\)

Hence, Loss of kinetic energy due to the direct impact of two bodies depends on mass of those two bodies and initial velocity of the two bodies 

Explanation:

Types of collision:

Perfectly elastic collision: A perfectly elastic collision is defined as one in which there is no loss of kinetic energy and the momentum of the system is conserved in the collision.

Inelastic collision: A inelastic collision is defined as one in which there is a loss of kinetic energy and the momentum of the system is conserved in the collision.

The coefficient of restitution is the ratio of relative velocity after impact to the relative velocity before impact. 

Coefficient of restitution (e)

\({\rm{e}} = \frac{{{\rm{Relative\;velocity\;after\;collision}}}}{{{\rm{Relative\;velocity\;before\;collision}}}} = \frac{{{{\rm{v}}_2} - {{\rm{v}}_1}}}{{{{\rm{u}}_1} - {{\rm{u}}_2}}}\)

• For perfectly elastic collision, e = 1
• For inelastic collision, 0 < e < 1
• For a perfectly inelastic collision, e = 0

Explanation:

This is the case of plastic impact as both balls stick after the collision. 

Plastic impact/ Perfectly inelastic impact

• The two bodies move together with common velocity after the impact
• Momentum is conserved, i.e. m1u1 + m2u2 = (m1 + m2) v₀
• where v_o is common velocity after impact.
• There is a loss in Kinetic energy after the impact.

Let, the two perfectly elastic spherical bodies are A and B

Mass both the bodies is the same that is m, the initial velocity of object A  is u_A, the initial velocity of body B is u_B, v₀ is common velocity after impact.

where u_A = v and u_B = 0.

Moment of conservation:

muA + muB = (m + m) v₀

m(v) =  2m × v₀

∴ \(v_0 ={v\over 2}\)

Additional Information

Newton's law of collision: Coefficient of Restitution.

• Considering two bodies A and B of mass m1 and m2 respectively. Let, these bodies moving with respective velocities u1 and u2 before impact. The impact will take place only if u1 is greater than u2.
• The velocity of approach = (u1 - u2)
• After a short period of constant, the bodies will separate and will start moving with velocities v1 and v2 respectively. The separation will occur only when V2 is greater than V1.
• The velocity of separation = (v2 - v1) 
• When two moving bodies collide with each other, their velocity of separation bears constant to their velocity of approach.

Mathematically: (v2 - v1) = e (u1 - u2)

Where e is a coefficient of restitution.

\(e={v_2-v_1\over u_1-u_2}\)

The coefficient of restitution is a parameter that indicates the energy loss during an impact.

The value of e lies between 0 and 1.​

Properties of different types of collision are given in the table below:

Concept:

The energy dissipation during impact is described by the term coefficient of restitution, which is a scalar quantity.

e = velocity of separation / velocity of approach

e = (v₁ - v₂) / (u₂ - u₁)

where, v = velocity of the body after impact, u = velocity of the body before impact

• For perfectly elastic collision, e = 1
• For inelastic collision, e < 1
• For perfectly inelastic collision, e = 0

Note:

When two bodies collide, they exchange energy. The body with higher initial energy transfers some of it to the one with lower energy. Hence, the body with initially higher velocity will have lower velocity after the impact.

Concept:

When the ball is dropped from some height (h) the velocity with which it impacts the surface is given by \(v = \sqrt {2gh} \)

Coefficient of restitution = \(\frac{{Velocity\;of\;Separation}}{{Velocity\;of\;Approach}} = \frac{{\left| {\left( {{v_2} - {v_1}} \right)} \right|}}{{\left| {\left( {{u_1} - \;{u_2}} \right)} \right|}}\)

u1 and u2 are the initial velocities of the ball and the surface respectively before impact

v1 and v2 are the velocities of the ball and the surface respectively after the impact

Calculation:

Before impact, the velocity is given by \({u_1} = \sqrt {2gh}\)

The initial and final velocity of the surface will be Zero

After the first impact \(e = \frac{{\left| {\left( {0 - {v_1}} \right)} \right|}}{{\left| {({u_1} - 0)} \right|}} = \frac{{{v_1}}}{{{u_1}}}\)

\(e = \frac{{\sqrt {2g{h_1}} }}{{\sqrt {2gh} }} = \sqrt {\frac{{{h_1}}}{h}}\)

Calculation:

Given:

h₁ = 1 m, h = 2.25 m

\(e=\sqrt{\frac{1}{2.25}}=0.66\)