Multiple Choice Questions on Elastic and Inelastic Collisions

Today we will discuss two questions on collisions. The first question pertains to elastic collision in one dimension and the second question pertains to inelastic collision in two dimensions:

(1) A particle A of mass m moving along the positive x-direction with kinetic energy K suffers an elastic head-on collision with a stationary particle B of mass M. After the collision the particle A moves along the negative x-direction with kinetic energy K/9. What is the mass of particle B?

(a) 9 m

(b) 6 m

(c) 3 m

(d) 2 m

(e) m/3

Since the kinetic energy of A after collision is one-nineth of its initial kinetic energy, the momentum of A after collision is one-third of its initial momentum.

Since the momentum is to be conserved, we have

p = p’ – p/3 where p is initial momentum of A and p’ is the momentum of B after the collision.

[The final momentum of particle A is negative since its direction is reversed].

Therefore, p’ = 4p/3

The kinetic energy gained by particle B due to the collision is p’²/2M where M is the mass of particle B.

The kinetic energy lost by particle A due to the collision is (8/9)×p²/2m.

[Note that the initial kinetic energy of particle A is p²/2m and its final kinetic energy is (1/9) p²/2m].

Since the kinetic energy too is conserved in elastic collisions, the kinetic energy gained by particle B is equal to the kinetic energy lost by particle A. Therefore, we have

p’²/2M = (8/9) p²/2m

Substituting for p’ = 4p/3, we have

(16/9) (p²/2M) = (8/9) p²/2m from which M = 2 m.

(2) A particle of mass m moving westwards with speed v collides with an identical particle moving northwards with speed v. If they stick together after the collision, what is their common speed after the collision?

(a) 4v

(b) 2√2 v

(c) 2v

(d) √2 v

(e) v/√2

The total momentum before collision is the vector sum of the initial momenta mv and mv at right angles (since one is westwards and the other is northwards) and is equal to √2 mv acting along the north-west direction (Fig.).

The combined mass is 2m. If V is the common speed, the total momentum after collision is 2m V, acting along the north-west direction.

Since the momentum is conserved, we have

2m V = √2 mv so that V = v/√2

IIT-JEE 2009 Multiple Choice Question (Single Answer Type) on Elastic Collision

The following question which appeared in IIT-JEE 2009 question paper is based on the principle that in an elastic collision between two particles of the same mass, the velocities get interchanged: 

Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v respectively as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A?

(a) 4

(b) 3

(c) 2

(d) 1    

Particle P₁, after traversing one third (AB) of the circular track in the anticlockwise direction, will collide with particle P₂ at position B. This collision occurs when the particle P₂  has traversed two thirds (ACB) of the circular path (since the speed of P₂ is twice that of P₁). 

         Since the collision is elastic and the particles are of equal masses, their velocities are interchanged. P₁ now travels in the clockwise direction with speed 2v and P₂ travels in the anticlockwise direction with speed v. The second collision takes place at position C. (BAC is two thirds of the circular path while BC is one third of the path). The second collision at C results in the reversal of the velocities and P₁ now travels in the anticlockwise direction with speed v where as P₂ travels in the clockwise direction with speed 2v. Therefore, the third collision will occur at A.

Since the third collision at A is not to be counted, the answer is 2 [Option (c)].

*    *    *    *    *    *    *    *    *    *    *    *   *    *    *    *    *

Now suppose we modify the above question as follows:

Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular frictionless track. Their speeds are v and 3v respectively as shown in the figure. After making how many elastic collisions, other than that at A, these two particles will again reach the point A ?

(a) 4

(b) 3

(c) 2

(d) 1

[You may use the figure shown with the question above, replacing the velocity 2v with 3v]

You can easily arrive at the answer which is 3 [Option (b)].

Angular Momentum- Two Multiple Choice Questions

How strange is the lot of us mortals! Each of us is here for a brief sojourn; for what purpose we know not, though some times sense it. But we know from daily life that we exist for other people first of all for whose smiles and well-being our happiness depends.

– Albert Einstein

(1) A thin uniform wooden rod AB of length L and mass M is hinged without friction at the end B. A small block of clay of mass m moving horizontally with velocity v srikes the end B of the rod and gets stuck to it. The angular velocity of the system about A just after the collision is

(a) mv/(mL + ML)

(b) 3mv/(mL + ML)

(c) (mv+ ML)/ 3mL

(d) mv/(3mL + ML)

(e) 3mv/(3mL + ML)

The angular momentum of the system about the point A just before collision is the angular momentum of the block of clay which is equal to mvL. (Note that the perpendicular distance of the line of action of the linear momentum (mv) of the block of clay from the point A is L).

The angular momentum of the system about the point A just after the collision is Iω where I is the total moment of inertia of the rod and clay and ω is the angular velocity of the system immediately after the collision.

We have I = ML²/3 + mL²

Equating the angular momenta before and after collision, we have

mvL = (ML²/3 + mL²) ω

Therefore ω = 3mv/(3mL + ML).

(2) A particle is projected at an angle of 60º with the horizontal with linear momentum of magnitude p. The horizontal range of this projectile is R. Just before the projectile strikes the ground at A, what is the magnitude of its angular momentum about an axis perpendicular the plane of motion and passing through the point of projection? Neglect air resistance.

(a) pR

(b) pR/2

(c) (√3) pR /2

(d) (√3) pR

(e) Zero

This is a very simple question. The magnitude of the linear momentum of the projectile just before it strikes the ground will be equal to p. The magnitude of the angular momentum at the moment will be p×Lever arm = p×ON = p×R sin 60º = (√3) pR /2.

Multiple Choice Questions (MCQ) on Elastic Collision

The figure shows three identical blocks A, B and C (each of mass m) on a smooth horizontal surface. B and C which are initially at rest, are connectd by a spring of negligible mass and force constant K. The block A is initially moving with velocity ‘v’ along the line joining B and C and hence it collides with B elastically.

Now, answer the following questions related to the system:

(1) After the collision the velocity of the block A is

(a) v/3

(b) – v

(c) – v/3

(d) v/2

(e) zero

At the moment the block A collides with the block B, you need consider these two blocks only. In one dimensional elastic collision between two bodies of the same mass, if one body is initially at rest, the moving body comes to rest and the body initially at rest moves with the velocity of the moving body. After the collision, the velocity of A is therefore zero.

[Generally, if the moving block 1 has mass m₁ and it moves with initial velocity v_1i towards the stationary block 2 of mass m₂, the final velocities of blocks 1 and 2 are given by

v_1f = v_1i(m₁ – m₂) /(m₁ + m₂) and

v_2f = 2 m₁v_1i /(m₁ + m₂)

The above relations can be easily obtained from momentum and kinetic energy conservation equations.

(2) When the compression of the spring is maximum, the velocity of block B is

(a) v

(b) – v

(c) – v/3

(d) v/2

(e) zero

On receiving the entire momentum (mv) of the block A, the block B moves towards the block C, compressing the spring. When the compression of the spring is maximum, both B and C move with the same velocity (v’). Since B and C have the same mass (m) and momentum has to be conserved, we have

mv = (m + m) v’

Therefore v’ = v/2.

(3) When the compression of the spring is maximum, the kinetic energy of the system is

(a) (½) mv²

(b) (1/4) mv²

(c) Zero

(d) mv²

(e) (1/8) mv²

The block A has come to rest after colliding with block B. When the compression of the spring is maximum, both B and C are moving with the same velocity v/2 as shown above. Therefore, the kinetic energy of the system under this condition is 2×(1/2)m(v/2)² = (1/4) mv². [Note that the spring has negligible kinetic energy since its mass is negligible].

(4) The maximum compression of the spring is

(a) √(mv²/2K)

(b) √(mv/2K)

(c) mv/2K

(d) mv²/2K

(e) √(2mv²/K)

The total energy of the system is equal to the initial kinetic energy (½ )mv² of the colliding block A. Therfore we have

(½)mv² = KE of A + KE of B + KE of C + PE of spring. The KE of A is zero after the collision. As shown above, the total KE of B and C is (1/4) mv² when the compression of the spring is maximum. Therfore, the potential energy (PE) of the spring at maximum compression is given by the equation,

(½)mv² = (1/4) mv² + (1/2)Kx² where x is the maximum compression of the spring.

This gives (1/2)Kx² = (1/4) mv² from which x = √(mv²/2K).

You will find similar questions (with solution) from different branches of physics at apphysicsresources.blogspot.com

IIT-JEE 2007 Assertion-Reason Type MCQ on Elastic Collisions

The following simple Assertion-Reason Type multiple choice question which appeared in IIT-JEE 2007 question paper is quite simple and is meant for checking whether you have mastered the fundamentals of elastic and inelastic collisions:

STATEMENT-1

In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision

because

Statement-2

In an elastic collision, the linear momentum of the system is conserved.

(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct

explanation for Statement-1

(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct

explanation for Statement-1

(c) Statement-1 is True, Statement-2 is False

(d) Statement-1 is False, Statement-2 is True

The important points you have to remember here are:

(i) momentum is conserved in elastic as well as inelastic collisions

(i) in elastic collisions, kinetic energy also is conserved.

[Remember, we are talking of kinetic energy. Note that the total energy is conserved in all collisions in accordance with the law of conservation of energy].

In an elastic collision, the relative speed of the bodies after collision is equal to the relative speed before the collision and this is a consequence of the conservation of kinetic energy.

Therefore, statement-1 is true. Statement-2 also is true, but it is not the reason for statement-1. So, the correct option is (b).

[Note that the ratio of the relative speed of the bodies after collision to the relative speed before collision is called the coefficient of restitution, e. The value of ‘e’ lies between 0 and 1. For perfectly elastic collisions, e =1 and for perfectly inelastic collisions, e = 0]

Two Multiple Choice Questions on Elastic Collision

You should remember that momentum and kinetic energy are conserved in elastic collisions where as momentum alone is conserved in inelastic collisions. Let us consider two questions (MCQ) involving elastic collision in one dimension:

(1) A system consisting of two identical blocks A and B, each of mass ‘m’, connected by a light spring of force constant ‘k’ is resting on a smooth horizontal surface. A third identical block C of mass ‘m’ moving with a velocity ‘v₀’along the direction of the line joining A and B collides elastically with A and compresses the spring. The maximum compression of the spring is

(a) √(mv₀/k) (b) √(mv₀/2k) (c) √(m/v₀k) (d) v₀√(m/k) (e) v₀√(m/2k)

At the instant of collision, you need consider the two identical colliding masses C and A only. As the collision is elastic, the entire momentum and kinetic energy of C are transferred to A and C comes to rest. The block A then moves towards B, compressing the spring. When the compression is maximum, both A and B move with the same velocity.

Since the momentum is always conserved,

mv₀ = (m + m)v where ‘v’ is the common velocity with which the masses A and B move. Therefore, v = v₀/2.

The entire kinetic energy of C is transferred to A, which then compresses the spring and pushes B so that we have

½ mv₀² = ½ (m + m)(v₀/2)² + ½ kx², where ‘x’ is the maximum compression.

This gives x = v√(m/2k).

(2) An α-particle of mass ‘m’ suffers a one-dimensional elastic collision with a nucleus of unknown mass at rest. After the collision the α-particle is scattered directly backwards, losing 75% of its kinetic energy. Then, the mass of the nucleus is

(a) m (b) 2m (c) 3m (d) (3/2)m (e) 5m

This MCQ appeared in the Kerala Engineering Entrance 2003 question paper.

The total momentum of the system is equal to the initial momentum p₁ of the α-particle. Equating the total initial and final momenta,

p₁ = P – p₂ where P and p₂ are respectively the final momenta of the nucleus and the α-particle. (The negative sign for p₂ is because of the backward motion of the α-particle after the collision).

But the magnitude of p₂ is p₁/2 since the final kinetic energy of the α-particle is ¼ of its initial kinetic energy. [Remember that K.E. = p²/2m and hence the momentum p is directly proportional to the square root of the kinetic energy].

Therefore, P = p₁ + (p₁/2) = (3/2) p₁.

Since the collision is elastic, kinetic energy also is conserved.

Therefore, K.E. lost by the α-particle = K.E gained by the nucleus so that

P²/2M = (3/4) p₁² /2m where ‘M’ is the mass of the nucleus.

Substituting for P, [(3/2)p₁]²/ 2M = [(3/4)p₁²]/2m

This gives M = 3m.