Collision of Bodies MCQ Quiz in தமிழ் - Objective Question with Answer for Collision of Bodies - இலவச PDF ஐப் பதிவிறக்கவும்
Last updated on Mar 10, 2025
Latest Collision of Bodies MCQ Objective Questions
Top Collision of Bodies MCQ Objective Questions
Collision of Bodies Question 1:
According to the law of collision of elastic bodies, the value of Coefficient of Restitution (e) for the two perfect elastic bodies is
Answer (Detailed Solution Below)
Collision of Bodies Question 1 Detailed Solution
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
Collision of Bodies Question 2:
Which of the following statements are true regarding collisions?
Answer (Detailed Solution Below)
Collision of Bodies Question 2 Detailed Solution
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.
Collision of Bodies Question 3:
If the two bodies, before impact, are not moving along the line of impact, the collision is called
Answer (Detailed Solution Below)
Collision of Bodies Question 3 Detailed Solution
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.
Collision of Bodies Question 4:
A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up to a height d/2. Neglecting the subsequent motion and air resistance, its velocity (v) varies with height (h) above the ground as
Answer (Detailed Solution Below)
Collision of Bodies Question 4 Detailed Solution
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.
As the height decreases, the negative velocity increase and becomes maximum at h = 0 and after striking jumps to some height till the velocity becomes zero.Collision of Bodies Question 5:
Collision is said to be ________ if both the particles stick together after collision and move with same velocity.
Answer (Detailed Solution Below)
Collision of Bodies Question 5 Detailed Solution
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.
- Based on this law of conservation this classification is made, and the laws of collision state:
- 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.
Collision of Bodies Question 6:
The loss of kinetic energy due to the direct impact of two bodies depends on,
Answer (Detailed Solution Below)
Collision of Bodies Question 6 Detailed Solution
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:
\({\rm{\Delta }}E = \frac{{{m_1}{m_2}}}{{2\left( {{m_1} + {m_2}} \right)}}{\left( {{v_1} - {v_2}} \right)^2}(1-e^2)\)
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
Collision of Bodies Question 7:
The co-efficient of restitution of a perfectly plastic impact is
Answer (Detailed Solution Below)
Collision of Bodies Question 7 Detailed Solution
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) =\(\frac{Relative\;velocity\;after\;collosion}{Relative\;velocity\;before\;collosion}=\frac{v_B-v_A}{u_A-u_B}\)
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:
Types of Collision |
Linear momentum |
Total energy |
Kinetic energy |
Coefficient of restitution |
Perfectly elastic collision |
Conserved |
Conserved |
Conserved |
e = 1 |
Inelastic collision |
Conserved |
Conserved |
Not-Conserved |
0 < e < 1 |
Perfectly inelastic collision |
Conserved |
Conserved |
Not-Conserved |
e = 0 |
Collision of Bodies Question 8:
For inelastic bodies, the coefficient of restitution is ______.
Answer (Detailed Solution Below)
Collision of Bodies Question 8 Detailed Solution
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
Hence most appropriate answer is 0.8 here.
Collision of Bodies Question 9:
A railway wagon A of mass 10000 kg collides with another identical wagon B as shown in the figure. If A is moving at 5 m/s and B is at rest at the time of collision, the maximum compression in the spring S with a spring constant of 2 MN/m will be
Answer (Detailed Solution Below)
Collision of Bodies Question 9 Detailed Solution
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:
\( \frac{1}{2} × 10000 × {5^2} = \frac{1}{2} × K{x^2}\)
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
Collision of Bodies Question 10:
A ball of mass 0.5 kg moving at 10 m/s collides with another stationary ball of mass 1 kg and comes to a complete stop. What is the velocity of the second ball post-collision? (Assume an elastic collision)
Answer (Detailed Solution Below)
Collision of Bodies Question 10 Detailed Solution
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.