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

Answer : 3

Solution :

In elastic collision, there is a loss in kinetic energy. However, momentum is conserved if there is no external force acting on the system.

Concept:

• Conservation of Linear Momentum:
• In an inelastic collision, where two particles stick together after the collision, the total momentum before the collision is equal to the total momentum after the collision.
• The formula for linear momentum is given by: p = mv, where:
  • p: Momentum (kg·m/s)
  • m: Mass of the particle (kg)
  • v: Velocity of the particle (m/s)
• For this problem, since the two particles stick together, the momentum before the collision is equal to the momentum after the collision.

Calculation:

Given,

Mass of the first particle, m = m

Velocity of the first particle before the collision, v₁ = v

Mass of the second particle, m₂ = 2m

Velocity of the second particle before the collision, v₂ = 0 (since it's stationary)

Using the conservation of linear momentum:

Initial momentum = Final momentum

m × v + 2m × 0 = (m + 2m) × v′

mv = 3m × v′

⇒ v′ = v / 3

∴ The velocity of the combined particles after the collision is v / 3.

Calculation:
Given: Block A has a mass of 4 kg, and block B has a mass of 5 kg. Block B rests on a smooth horizontal table, and the minimum force required to move both blocks together is 12 N.

Let the coefficient of static friction between block A and block B be μ. For the blocks to move together, the frictional force between the two blocks must provide the necessary force to move both blocks as a single unit.

The frictional force, f, between block A and block B is given by:

f = μ × N, where N is the normal force between the two blocks. Since block A is placed on block B, the normal force is equal to the weight of block A, which is:

N = m_A × g = 4 × 10 = 40 N

Therefore, the maximum frictional force is:

f = μ × 40 N

Now, the force required to move both blocks together is:

F = (m_A + m_B) × a, where a is the acceleration of the blocks.

We know that the minimum force required to move the blocks together is 12 N, and this force must overcome the friction between the blocks. Thus, the frictional force must be equal to the applied force for both blocks to move together:

12 N = μ × 40 N

Solving for μ:

μ = 12 / 40 = 0.3

Now, the maximum force that can be applied to block B without sliding is:

F_max = (m_A + m_B) × g × μ = (4 + 5) × 10 × 0.3

F_max = 9 × 10 × 0.3 = 27 N

Thus, the correct option is Option 3: 27 N

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:

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

\(\Rightarrow V = \frac{m_1u_1}{m_1+m_2}\)  

The energy transferred to m2, \(KE_2=\frac 12m_2v_2^2=\frac12 m_2\frac{m_1^2u_1^2}{(m_1+m_2)^2}\)

\(KE_2=\frac{m_1^2m_2u_1^2}{2(m_1+m_2)^2}=\frac{m_1^2m_2u_1^2}{2(m_1-m_2)^2+8m_1m_2}\)

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) = − \( {1\over 3}\) (-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

\(\Rightarrow V = \frac{m_1u_1}{m_1+m_2}\)  

• By the law of conservation of kinetic energy

\(\Rightarrow \frac{1}{2}m_1u^2_1=\frac{1}{2}(m_1+m_2)V^2\)

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

\(\frac{KE_f}{KE_i}=\frac{\frac{1}{2}(m_1+m_2)V^2}{\frac{1}{2}mu^2_1}=\frac{\frac{1}{2}(m_1+m_2)[\frac{m_1u_1}{m_1+m_2}]^2}{\frac{1}{2}mu^2_1}=\frac{m_1}{m_1+m_2}\)

• The fractional loss in the kinetic energy is

\(\Rightarrow \frac{KE_i-KE_f}{KE_i}=\frac{KE_i([1-\frac{KE_f}{KE_i}])}{KE_i}=\frac{KE_i([1-\frac{m_1}{m_1+m_2}])}{KE_i}=\frac{m_2}{m_1+m_2}\)