Collision of Bodies MCQ Quiz in தமிழ் - Objective Question with Answer for Collision of Bodies - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

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

$e=\frac{Velocityofseparation}{Velocityofapproach}$

$e=\frac{v_1-v_2}{u_2-u_1}$

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:

Coefficient of restitution (e) for various types of collisions is given by

e = 1 → Elastic impact (Option 1)

e = 0 → Plastic impact

e < 1 → Inelastic impact (Option 2 is wrong)

e > 1 → Super elastic impact

Total energy is conserved in both elastic and inelastic collisions (Option 3)

Whereas K.E is conserved in Elastic collision and not conserved in inelastic collisions.

Concept:

• If the two bodies, before impact, are not moving along the line of impact, the collision is known as indirect impact or oblique impact.
• In figure (a) shows the two bodied A nad B, moving with a velocity of u₁ and u₂ in different directions.
• The velocity of body A makes an angle of θ₁ with horizontal direction whereas the velocity of body B makes an angle of θ₂ with the horizontal direction.
• As the two bodies are moving in different directions, the collision between two bodies takes place as shown in figure (b).
• The point C is the point of contact and the line joining the centers O1O2 with point C is known as the line of impact.

The collision between two bodies is known as the direct impact if the two bodies before impact, are moving along the line of impact.

Concept:

From the equation of motion,

v² = u² + 2ah

Since the ball is dropped, so u = 0

So, the graph between velocity and height will be parabolic.

Explanation:

Coefficient of restitution (e):

Coefficient of restitution or coefficient of the resilience of a collision is defined as the ratio of the relative velocity of separation after the collision to the relative velocity of the approach before the collision.

Coefficient of restitution (e) =$\frac{Relativevelocityaftercollosion}{Relativevelocitybeforecollosion}=\frac{v_B-v_A}{u_A-u_B}$

A perfectly inelastic collision also called a perfectly plastic collision is a limiting case of inelastic collision in which the two bodies stick together after impact.

Since both bodies stick together in a perfectly plastic collision. Therefore vA = vB, Thus e = 0

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

CONCEPT:

• Collison’s are generally classified into two types elastic and inelastic collisions
  • Based on this law of conservation this classification is made, and the laws of collision state:
    • Momentum is conserved in all collisions.
    • In an elastic collision, kinetic energy is also conserved.
    • In an inelastic collision, kinetic energy is not conserved. In a perfectly inelastic collision, objects stick together after the collision.
• 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.
  • But inelastic Collison obeys the law of conservation of linear momentum.
  • In a perfectly inelastic collision, both the particles stick to each other and move together.

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  and $v_1^{,},v_2^{,}$ are final velocities of the colliding bodies.

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

$\mathrm{\Delta }E=\frac{m_1{m}_2}{2\left(m_1+m_2\right)}{\left(v_1-v_2\right)}^2(1-e^2)$

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 

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:

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)

$\mathrm{e}=\frac{\mathrm{R}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{a}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{c}\mathrm{o}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{o}\mathrm{n}}{\mathrm{R}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{b}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{o}\mathrm{n}}=\frac{{\mathrm{v}}_2-{\mathrm{v}}_1}{{\mathrm{u}}_1-{\mathrm{u}}_2}$

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

Hence most appropriate answer is 0.8 here.

Concept:

For maximum compression perfect inelastic collision. For maximum compression the total kinetic energy of the wagon A is converted into potential energy of spring

$\frac{1}{2}$× m × v² = $\frac{1}{2}$× k × x²

Calculation:

Given:

$\frac{1}{2}\times 10000\times 5^2=\frac{1}{2}\times K{x}^2$

10000 × 25 = 2 × 10⁶ × x²

Therefore, $x=\frac{5}{10\sqrt{2}}=\frac{5\sqrt{2}}{10\times 2}$ 

$=\frac{\sqrt{2}}{4}m=\frac{\sqrt{2}}{4}\times 100$ = 35.35 cm