Collisions MCQ Quiz - Objective Question with Answer for Collisions - Download Free PDF
Last updated on May 29, 2025
Latest Collisions MCQ Objective Questions
Collisions Question 1:
Consider the following statements A and B. Identify the correct choice in the given answers.
A. In an inelastic collision, there is no loss in kinetic energy during collision.
B. During a collision, the linear momentum of the entire system of particles is conserved if there is no external force acting on the system.
Answer (Detailed Solution Below)
Collisions Question 1 Detailed Solution
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.
Collisions Question 2:
A particle of mass m moving with velocity v collides with a stationary particle of mass 2m. After collision, they stick together and continue to move together with velocity
Answer (Detailed Solution Below)
Collisions Question 2 Detailed Solution
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.
Collisions Question 3:
A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a smooth horizontal table. If the minimum force that can be applied on A so that both the blocks move together is 12 N, the maximum force that can be applied to B for the blocks to move together will be
Answer (Detailed Solution Below)
Collisions Question 3 Detailed Solution
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 = mA × 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 = (mA + mB) × 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:
Fmax = (mA + mB) × g × μ = (4 + 5) × 10 × 0.3
Fmax = 9 × 10 × 0.3 = 27 N
Thus, the correct option is Option 3: 27 N
Collisions Question 4:
Two solid rubber balls A and B having masses 200 & 400 gm respectively are moving in opposite direction with velocity of A equal to 0.3 m/sec. After collision the two balls come to rest when the velocity of B is
Answer (Detailed Solution Below)
Collisions Question 4 Detailed Solution
Calcultion:
Initial momentum of Ball A = 0.2 × 0.3 = 0.06 kg·m/s
Initial momentum of Ball B = 0.4 × vB
Total initial momentum = 0.06 + 0.4 × vB
According to conservation of momentum:
0.06 + 0.4 × vB = 0
0.4 × vB = -0.06
vB = -0.06 / 0.4
vB = -0.15 m/sec
Collisions Question 5:
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) :
Three identical spheres of same mass undergo one dimensional motion as shown in figure with initial velocities vA = 5 m/s, vB = 2 m/s, vC = 4 m/s If we wait sufficiently long for elastic collision to happen, then vA = 4 m/s, vB = 2 m/s, vC = 5 m/s will be the final velocities.
Reason (R) : In an elastic collision between identical masses, two objects exchange their velocities.
In the light of the above statements, choose the correct answer from the options given below :
Answer (Detailed Solution Below)
Collisions Question 5 Detailed Solution
Top Collisions MCQ Objective Questions
A sphere of mass 2kg strikes another sphere of mass 3 kg at rest with a velocity of 5 m/s. if they move together after collision. What is their common velocity?
Answer (Detailed Solution Below)
Collisions Question 6 Detailed Solution
Download Solution PDFConcept:
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.
When two bodies collide elastically then the quantity conserved is
Answer (Detailed Solution Below)
Collisions Question 7 Detailed Solution
Download Solution PDFConcept:
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:
Elastic collision |
Inelastic collision |
The kinetic energy is conserved. |
The kinetic energy is not conserved. |
The momentum is conserved. |
The momentum is conserved. |
The total energy is conserved. |
The total energy is conserved. |
Forces involved during a collision are conservative. |
Some or all of the forces involved during a collision are non-conservative. |
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.
A 100 g sphere is moving at a speed of 20 m/s and collides with another. sphere of mass 50 g. If the second sphere was at rest prior to the collision and the first sphere comes at rest immediately after the collision, considering the collision to be elastic, the speed of the second sphere would be
Answer (Detailed Solution Below)
Collisions Question 8 Detailed Solution
Download Solution PDFThe 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*v1i + m2*v2i = m1*v1f + m2*v2f
- where:
- m1 and m2 are the masses of the first and second spheres, respectively,
- v1i and v2i are their initial velocities,
- v1f and v2f are their final velocities.
- m1*v1i + m2*v2i = m1*v1f + m2*v2f
- Since the second sphere is initially at rest (v2i = 0) and the first sphere comes to rest (v1f = 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*v2f
- Solving for v2f :
- v2f = 40m/s
- Therefore, the speed of the second sphere after the collision is 40 m/s.
In an elastic collision, the kinetic energy of the system ______.
Answer (Detailed Solution Below)
Collisions Question 9 Detailed Solution
Download Solution PDFCONCEPT:
- 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:
Elastic collision |
Inelastic collision |
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For elastic and inelastic collision which of the statement is CORRECT.
Answer (Detailed Solution Below)
Collisions Question 10 Detailed Solution
Download Solution PDFCONCEPT:
- 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.
Two solid rubber balls A and B having masses 200 & 400 gm respectively are moving in opposite direction with velocity of A equal to 0.3 m/sec. After collision the two balls come to rest when the velocity of B is
Answer (Detailed Solution Below)
Collisions Question 11 Detailed Solution
Download Solution PDFCalcultion:
Initial momentum of Ball A = 0.2 × 0.3 = 0.06 kg·m/s
Initial momentum of Ball B = 0.4 × vB
Total initial momentum = 0.06 + 0.4 × vB
According to conservation of momentum:
0.06 + 0.4 × vB = 0
0.4 × vB = -0.06
vB = -0.06 / 0.4
vB = -0.15 m/sec
In a perfectly inelastic direct collision maximum transfer of energy takes place if -
Answer (Detailed Solution Below)
Collisions Question 12 Detailed Solution
Download Solution PDFConcept:
- 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
According to the law of conservation of momentum, in an ideal closed system, when two objects collide:
Answer (Detailed Solution Below)
Collisions Question 13 Detailed Solution
Download Solution PDFExplanation:
- 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.
A ball is moving with a speed of 5 m/s and collides with another ball moving with a speed of 1 m/s in the opposite direction. What will be the speed of the second ball if after the collision the first ball comes to rest and the coefficient of restitution is \( {1\over 3}\)?
Answer (Detailed Solution Below)
Collisions Question 14 Detailed Solution
Download Solution PDFCONCEPT:
- 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.
A particle of mass m1 moving with u1 velocity collide with another particle of mass m2 which is initially in rest. If collision is perfectly inelastic then the fractional loss in the kinetic energy is
Answer (Detailed Solution Below)
Collisions Question 15 Detailed Solution
Download Solution PDFCONCEPT:
- 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
\(\Rightarrow \frac{KE_i-KE_f}{KE_i}=\frac{KE_i([1-\frac{KE_f}{KE_i}])}{KE_i}=\frac{KE_i([1-\frac{m_1}{m_1+m_2}])}{KE_i}=\frac{m_2}{m_1+m_2}\)