Collisions MCQ Quiz - Objective Question with Answer for Collisions - Download Free PDF

Explanation:

According to Newton's second law, Fext = dp/dt, in the absence of external forces (Fext = 0), the equation simplifies to dp/dt = 0. This means that the linear momentum p of the system is constant when the resultant of all external forces is zero.

The kinetic energy of the system can still change, as seen in inelastic collisions. Similarly, angular momentum may change if an external torque is applied, such as in the case of a couple acting on a rod. Angular momentum is conserved only when there is no external torque.

Thus, the correct answer is (A): The linear momentum of the system does not change in time.

Calculation:

In a perfectly elastic collision:

The relative speed after collision equals the relative speed before collision because the coefficient of restitution e = 1.

This relationship comes from the definition of elasticity, not directly from momentum conservation.

Linear momentum is also conserved in all types of collisions (elastic and inelastic), provided no external force acts.

Hence, both statements are true, but the second is not a correct explanation of the first.

Calculation:

As the ball falls, its gravitational potential energy decreases and kinetic energy increases. When it contacts the surface, the kinetic energy is stored as elastic potential energy (like a spring), forming a peak. After rebounding, the cycle repeats. The variation in potential energy with time is thus smooth and periodic, with parabolic peaks at collision points.

Correct Option: (B)

Calculation:

From the conservation of momentum:

2m v₁ = 2m v_1f cos θ + 2m v_2f cos ϕ     (i)

2m v_1f sin θ = m v_2f sin ϕ     (ii)

Using conservation of kinetic energy:

1/2 (2m) v₁² + 1/2 m (0)² = 1/2 (2m) v_1f² + 1/2 m v_2f²     (iii)

⇒ 2 v₁² = 2 v_1f² + v_2f²     (iv)

From equations (i), (ii), and (iii), we get:

3 v_1f² − 4 v₁ v_1f cos θ + v_2f² = 0

⇒ (−4 v₁ cos θ)² − 4(3)(v₁²) ≥ 0

cos² θ ≥ 3/4

⇒ cos θ ≥ √3/2

⇒ θ = π / 6

Therefore, the correct option is D: π / 6.

Concept:

The energy dissipation during impact is called by the term, coefficient of restitution, 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 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 lower energy. Hence, the body which had the higher velocity initially will now have the lower velocity.

Concept:

Momentum:

momentum is the product of the mass and velocity of an object. It is a vector quantity, possessing a magnitude and a direction.

• The unit of momentum (P) is kg m/s.
• Dimension: [MLT^-1]

Law of conservation of Momentum: 

A conservation law states that the total linear momentum of a closed system remains constant through time, regardless of other possible changes within the system.

P₁ = P₂

m₁ v₁ = m₂ v₂

Where, P₁ = initial momentum of system, P₂ = final momentum of system, m₁ = mass of first object, v₁ = velocity of first object, m₂ = mass of second object and v₂ = velocity of second object.

Calculation:

Given:  m₁ = 2 kg    m₂ = 3 kg     u₁ = 5 m/s        u₂ ​=  0 m/s

Let the common velocity of the combined body be V m/s

Mass of combined body      M = 2 + 3 = 5 kg

Applying conservation of momentum:          

m₁ v₁ + m₂ v₂ = M V

⇒ (2 × 5) + (3 × 0) = 5 V

⇒ 10 + 0 = 5 V

⇒ V = 2 m/s

Hence the combined velocity of both the spheres is 2 m/s.

Concept:

Collision: 

A collision is said to occur between two objects, either if they physically collide against each other or if the path of one object is affected by the force exerted by the other objects.

Characteristic of elastic and inelastic collision:

Explanation:

From the above, it is clear that when two bodies collide elastically then both kinetic energy and momentum are conserved. Therefore option 4 is correct.

The correct answer is 40 m/s.

Key Points

• In an elastic collision between two spheres, both momentum and kinetic energy are conserved.
• The conservation of momentum can be expressed as:
  • m₁*v_1i ​ + m₂*v_2i  = m₁*v_1f ​ + m₂*v_2f ​
    • where:
    • m₁ and m₂ ​ are the masses of the first and second spheres, respectively,
    • v_1i and v_2i are their initial velocities,
    • v_{1f ​} and v_2f ​are their final velocities.
• Since the second sphere is initially at rest (v_2i ​= 0) and the first sphere comes to rest (v_1f​ = 0), the momentum conservation equation simplifies to: 
• m1*v1i ​ = m2*v2f
• Now, let's substitute the given values:
• 100 g*20 m/s = 50 g*v_2f
• Solving for v2f ​:
• v_2f ​ = 40m/s
• Therefore, the speed of the second sphere after the collision is 40 m/s.

CONCEPT:

• Collision: An event in which two or more bodies exert forces on each other for a very short time is called a collision.
  • It results in the exchange or transformation of energy.
• Elastic Collision: A collision that takes place between two objects in which no energy is lost.
• In the case of elastic collision, the kinetic energy and momentum both are conserved.

EXPLANATION:

• In the elastic collision, both momentum and kinetic energy are conserved.
• Since in the elastic collision kinetic energy of the system is conserved, it will remain constant.
• So the correct answer is option 3.

Additional Information

Characteristic of elastic and inelastic collision:

CONCEPT:

• Elastic collision: If the law of conservation of momentum and that of Conservation of energy (Only 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.

EXPLANATION:

• From the above explanation, we can conclude that in a perfectly elastic collision both energy and momentum are conserved.
• The term “elastic” indicates that there is no loss of energy in the form of heat or any such transformation during the collision.
• Thus, the only momentum is conserved in both elastic and inelastic collisions.

Calcultion:
Initial momentum of Ball A = 0.2 × 0.3 = 0.06 kg·m/s

Initial momentum of Ball B = 0.4 × v_B

Total initial momentum = 0.06 + 0.4 × v_B

According to conservation of momentum:

0.06 + 0.4 × v_B = 0

0.4 × v_B = -0.06

v_B = -0.06 / 0.4

v_B = -0.15 m/sec

Concept:

• A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision.
• An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision.

Explanation:

By the law of conservation of momentum:

⇒ m1u1 = (m1 + m2)V

$\Rightarrow V=\frac{m_1{u}_1}{m_1+m_2}$  

The energy transferred to m2, $K{E}_2=\frac{1}{2}{m}_2{v}_2^2=\frac{1}{2}{m}_2\frac{m_1^2{u}_1^2}{(m_1+m_2{)}^2}$

$K{E}_2=\frac{m_1^2{m}_2{u}_1^2}{2(m_1+m_2{)}^2}=\frac{m_1^2{m}_2{u}_1^2}{2(m_1-m_2{)}^2+8m_1{m}_2}$

KE₂ will be maximum when the denominator is minimum for which (m₁ - m₂)² = 0

m1 - m2 = 0

m1 = m2

Explanation:

• Perfectly elastic collision: The type of collision in which the total kinetic energy of the system remains constant before and after the collision is called elastic collision or perfectly elastic collision.
• Conservation of linear momentum: As there is no net external force on the system, so momentum before the collision and after the collision will remain same.

Linear Momentum before the collision (P₁) = Linear Momentum after the collision (P₂)

P₁ = m₁u₁ + m₂u₂

P₂ = m₁v₁ + m₂v₂

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

So according to conservation of linear momentum, after collision neither objects (considering both objects as system) loses nor gain any momentum.

CONCEPT:

• Collision: An event in which two or more bodies exert forces on each other for a very short time is called a collision.
  • It results in the exchange or transformation of energy.
• Coefficient of restitution: It is the ratio of the final relative velocity between two objects after they collide to the initial relative velocity between two objects before they collide.

Newton's law of restitution: Mathematically, it says

​v2 − v1= − e (u2 − u1)

where u1 is the initial velocity of the first ball, u2 is the initial velocity of the second ball, v1 is the final velocity of the first ball, v2 is the final velocity of the second ball, and e is the coefficient of restitution.

The coefficient of restitution always satisfies 0 ≤ e ≤ 1. 

CALCULATION:

Given that u1 = 5m/s; u₂ = -1m/s (-ve due to opposite direction); v₁ = 0 m/s and v2 = ? and e = 1/3

We know that ​v2 − v1= − e (u2 − u1)

​v2 − (0) = − $\frac{1}{3}$ (-1 − 5)

​v2 = −2 m/s (here -ve means opposite to initial direction)

So the correct answer is option 1.

Additional Information

• When e = 0, the balls remain in contact after the collision. (perfect inelastic collision)
• When e = 1, the collision is elastic.

CONCEPT:

• A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision.
• An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision.

EXPLANATION:

• By the law of conservation of momentum

⇒ m₁u₁ = (m₁ + m₂)V

$\Rightarrow V=\frac{m_1{u}_1}{m_1+m_2}$  

• By the law of conservation of kinetic energy

$\Rightarrow \frac{1}{2}{m}_1{u}_1^2=\frac{1}{2}(m_1+m_2)V^2$

• The ratio of kinetic energy before and after the collision is 

$\frac{K{E}_f}{K{E}_i}=\frac{\frac{1}{2}(m_1+m_2)V^2}{\frac{1}{2}m{u}_1^2}=\frac{\frac{1}{2}(m_1+m_2)[\frac{m_1{u}_1}{m_1+m_2}{]}^2}{\frac{1}{2}m{u}_1^2}=\frac{m_1}{m_1+m_2}$

• The fractional loss in the kinetic energy is

$\Rightarrow \frac{K{E}_i-K{E}_f}{K{E}_i}=\frac{K{E}_i([1-\frac{K{E}_f}{K{E}_i}])}{K{E}_i}=\frac{K{E}_i([1-\frac{m_1}{m_1+m_2}])}{K{E}_i}=\frac{m_2}{m_1+m_2}$