Physics 9702 Doubts | Help Page 25

Question 138: [Dynamics > Momentum]

Diagram shows two spherical masses approaching each other head-on at an equal speed u.

One has mass 2m and other has mass m.

Which diagram, showing situation after the collision, shows result of an elastic collision?

Reference: Past Exam Paper – November 2009 Paper 12 Q8 & November 2009 Paper 11

Q9

Solution 138:

Answer: A.

For an elastic collision,

Velocity of approach (before collision) = Velocity of separation (after collision)

{The above result can be obtained by considering that for elastic collision, both momentum

and kinetic energy is conserved. Momentum, p = mv. Kinetic energy = ½mv2. By equating

the sum of momentum before collision to that after collision and by equating the sum of KE

before collision to that after collision, 2 equations are obtained which can be simplified into

the above stated result: Velocity of approach (before collision) = Velocity of separation (after

collision). The proof will not be shown here}

{Approach means that the 2 sphere are coming towards each other and separation means that

they are moving away from each other}

Before collision, velocity of approach = u + u = 2u

Consider A:

Velocity of separation = (u/3) + (5u/3) = 6u/3 = 2u

Consider B:

Velocity of separation = (u/6) + (2u/3) = 5u/6

Consider C:

Both spheres are moving in the same direction. So, speed of separation is the difference in the

2 speed here/

Velocity of separation = (2u/3) – (u/6) = 3u/6 = 0.5u

Consider D:

Since the spheres stick together, they are not separating. They move together.

Velocity of separation = 0

Only answer A gives velocity of separation = 2u.

Question 139: [Dynamics > Inelastic collision]

Which quantities are conserved in inelastic collision?

Reference: Past Exam Paper – June 2008 Paper 1 Q17

Solution 139:

Answer: C.

The laws of conservation of energy and conservation of momentum state that, in a system, the

total energy and total momentum is always conserved.

Momentum in a system is always conserved in any collision. Linear momentum of the

momentum of a body in 1 dimension (linear motion).

In an inelastic collision, kinetic energy is NOT conserved. Some of the initial kinetic energy

is converted into other forms of energy. So, the total energy is still conserved

Question 162: [Dynamics > Elastic collision]

Two spheres travel along same line with velocities u1 and u2. They collide and after collision

their velocities are v1 and v2.

Which collision is not elastic?

Reference: Past Exam Paper – June 2013 Paper 13 Q11

Solution162:

Answer: A

For an elastic collision, the relative speed of approach is equal to the relative speed of

separation. (Note that approach means coming towards each other, and separation means

going away from each other.)

In this question, we need to identify which one is NOT elastic.

Let the sphere on the back be sphere 1 and the front sphere be sphere 2.

{Before proceeding with any calculations, you need to think how the spheres are moving

before and after collision from the signs (+ve or -ve) of the velocities.}

Consider A:

Before collision sphere 1 is moving forward (u1 = 2) while sphere 2 is moving backward (u2 =

-5). So, they are approaching.

Relative speed of approach = 2 + 5 = 7ms-1

After collision, both spheres 1 and 2 are moving backward, but sphere 1 is moving faster.

Relative speed of separation = 5 – 2 = 3ms-1

So, choice A is correct here

Consider B:

Before collision, both spheres are approaching.

Relative speed of approach = 3 + 3 = 6ms-1

After collision, sphere 1 is at rest and sphere 2 is moving forward.

Relative speed of separation = 6ms-1

Consider C:

Relative speed of approach = 3 – 2 = 5ms-1

Relative speed of separation = 6 – 1 = 5ms-1

Consider D:

Relative speed of approach = 5 – 2 = 3ms-1

Relative speed of separation = 6 – 3 = 3ms-1

Question 723: [Dynamics > Collision]

Particle of mass 2m and velocity v strikes a wall.

The particle rebounds along same path after colliding with the wall. The collision is inelastic.

What is a possible change in momentum of the ball during the collision?

A mv

B 2mv

C 3mv

D 4mv

Reference: Past Exam Paper – November 2010 Paper 12 Q9

Solution 723:

Answer: C.

Initial momentum of particle = (2m)v = 2mv

Initial kinetic energy of particle = ½ (2m) v2 = mv2

In an inelastic collision, energy is lost – so the kinetic energy of the particle must be less than

it was before the collision. But momentum is still conserved.

The wall is fixed (not moving), so its initial momentum is zero.

After colliding with the wall, the particle rebounds along same path. So, its direction has

changed.

Momentum is a vector quantity, so we need to consider its direction too.

Consider the direction towards the wall to be positive.

Initial momentum of particle = (2m)v = +2mv

The final momentum of the particle needs to be negative since its direction has changed.

If the change in momentum Δm = mv,

Final momentum of particle = +2mv – mv = +mv

The direction is still positive. This means that the particle is still moving towards the wall

after collision. We know, from the question, that this is incorrect.

If Δm = 2mv,

Final momentum of particle = +2mv – 2mv = 0

This means that the particle remains stationary after collision. This is incorrect since we know

that the particle moves in the opposite direction.

If Δm = 3mv,

Final momentum of particle = +2mv – 3mv = – mv

The particle moves away from the wall. Mass of the particle = 2m, so the velocity of the

particle is now –v/2 so that the momentum = (2m) (–v/2) = – mv

Final kinetic energy of particle = ½ (2m) (v/2)2 = mv2 / 4

The energy is less than before since this is an inelastic collision.

If Δm = 4mv,

Final momentum of particle = +2mv – 4mv = – 2mv

The particle moves away from the wall. Mass of the particle = 2m, so the velocity of the

particle is now –v so that the momentum = (2m) (–v) = – 2mv

Final kinetic energy of particle = ½ (2m) (v)2 = mv2

Here, the energy is conserved, which is incorrect for an inelastic collision

Question 772: [Dynamics > Collisions]

Diagram shows a particle P, travelling at speed v, about to collide with a stationary particle Q

of the same mass. The collision is perfectly elastic.

Which statement describes motion of P and of Q immediately after the collision?

A P rebounds with speed ½ v and Q acquires speed ½ v.

B P rebounds with speed v and Q remains stationary.

C P and Q both travel in the same direction with speed ½ v.

D P comes to a standstill and Q acquires speed v.

Reference: Past Exam Paper – June 2011 Paper 12 Q13

Solution 772:

Answer: D.

For a perfectly elastic collision, both momentum and energy are conserved.

From the law of conservation of momentum,

Sum of momentum before collision = Sum of collision after collision

Before collision,

Sum of momentum, p = mv + 0 = mv

Total kinetic energy = ½ mv2

After collision,

Consider choice A,

Sum of momentum, p = m(0.5v) + m(-0.5v) = 0.

[A is incorrect]

Consider choice B,

Sum of momentum, p = - mv

[B is incorrect]

Consider choice C,

Sum of momentum, p = m(0.5v) + m(0.5v) = mv

Total kinetic energy = ½ m (0.5v)2 + ½ m (0.5v)2 = 0.25mv2

Consider choice D,

Sum of momentum, p = m(0) + mv = mv

Total kinetic energy = ½ mv2

Both momentum and energy are conserved

[C is incorrect]

Question 138: [Dynamics > Momentum]

Diagram shows two spherical masses approaching each other head-on at an equal speed u.

One has mass 2m and other has mass m.

Which diagram, showing situation after the collision, shows result of an elastic collision?

Reference: Past Exam Paper – November 2009 Paper 12 Q8 & November 2009 Paper 11

Q9

Solution 138:

Answer: A.

For an elastic collision,

Velocity of approach (before collision) = Velocity of separation (after collision)

{The above result can be obtained by considering that for elastic collision, both momentum

and kinetic energy is conserved. Momentum, p = mv. Kinetic energy = ½mv2. By equating

the sum of momentum before collision to that after collision and by equating the sum of KE

before collision to that after collision, 2 equations are obtained which can be simplified into

the above stated result: Velocity of approach (before collision) = Velocity of separation (after

collision). The proof will not be shown here}

{Approach means that the 2 sphere are coming towards each other and separation means that

they are moving away from each other}

Before collision, velocity of approach = u + u = 2u

Consider A:

Velocity of separation = (u/3) + (5u/3) = 6u/3 = 2u

Consider B:

Velocity of separation = (u/6) + (2u/3) = 5u/6

Consider C:

Both spheres are moving in the same direction. So, speed of separation is the difference in the

2 speed here/

Velocity of separation = (2u/3) – (u/6) = 3u/6 = 0.5u

Consider D:

Since the spheres stick together, they are not separating. They move together.

Velocity of separation = 0

Only answer A gives velocity of separation = 2u.

Question 139: [Dynamics > Inelastic collision]

Which quantities are conserved in inelastic collision?

Reference: Past Exam Paper – June 2008 Paper 1 Q17

Solution 139:

Answer: C.

The laws of conservation of energy and conservation of momentum state that, in a system, the

total energy and total momentum is always conserved.

Momentum in a system is always conserved in any collision. Linear momentum of the

momentum of a body in 1 dimension (linear motion).

In an inelastic collision, kinetic energy is NOT conserved. Some of the initial kinetic energy

is converted into other forms of energy. So, the total energy is still conserved

Question 162: [Dynamics > Elastic collision]

Two spheres travel along same line with velocities u1 and u2. They collide and after collision

their velocities are v1 and v2.

Which collision is not elastic?

Reference: Past Exam Paper – June 2013 Paper 13 Q11

Solution162:

Answer: A

For an elastic collision, the relative speed of approach is equal to the relative speed of

separation. (Note that approach means coming towards each other, and separation means

going away from each other.)

In this question, we need to identify which one is NOT elastic.

Let the sphere on the back be sphere 1 and the front sphere be sphere 2.

{Before proceeding with any calculations, you need to think how the spheres are moving

before and after collision from the signs (+ve or -ve) of the velocities.}

Consider A:

Before collision sphere 1 is moving forward (u1 = 2) while sphere 2 is moving backward (u2 =

-5). So, they are approaching.

Relative speed of approach = 2 + 5 = 7ms-1

After collision, both spheres 1 and 2 are moving backward, but sphere 1 is moving faster.

Relative speed of separation = 5 – 2 = 3ms-1

So, choice A is correct here

Consider B:

Before collision, both spheres are approaching.

Relative speed of approach = 3 + 3 = 6ms-1

After collision, sphere 1 is at rest and sphere 2 is moving forward.

Relative speed of separation = 6ms-1

Consider C:

Relative speed of approach = 3 – 2 = 5ms-1

Relative speed of separation = 6 – 1 = 5ms-1

Consider D:

Relative speed of approach = 5 – 2 = 3ms-1

Relative speed of separation = 6 – 3 = 3ms-1

Question 723: [Dynamics > Collision]

Particle of mass 2m and velocity v strikes a wall.

The particle rebounds along same path after colliding with the wall. The collision is inelastic.

What is a possible change in momentum of the ball during the collision?

A mv

B 2mv

C 3mv

D 4mv

Reference: Past Exam Paper – November 2010 Paper 12 Q9

Solution 723:

Answer: C.

Initial momentum of particle = (2m)v = 2mv

Initial kinetic energy of particle = ½ (2m) v2 = mv2

In an inelastic collision, energy is lost – so the kinetic energy of the particle must be less than

it was before the collision. But momentum is still conserved.

The wall is fixed (not moving), so its initial momentum is zero.

After colliding with the wall, the particle rebounds along same path. So, its direction has

changed.

Momentum is a vector quantity, so we need to consider its direction too.

Consider the direction towards the wall to be positive.

Initial momentum of particle = (2m)v = +2mv

The final momentum of the particle needs to be negative since its direction has changed.

If the change in momentum Δm = mv,

Final momentum of particle = +2mv – mv = +mv

The direction is still positive. This means that the particle is still moving towards the wall

after collision. We know, from the question, that this is incorrect.

If Δm = 2mv,

Final momentum of particle = +2mv – 2mv = 0

This means that the particle remains stationary after collision. This is incorrect since we know

that the particle moves in the opposite direction.

If Δm = 3mv,

Final momentum of particle = +2mv – 3mv = – mv

The particle moves away from the wall. Mass of the particle = 2m, so the velocity of the

particle is now –v/2 so that the momentum = (2m) (–v/2) = – mv

Final kinetic energy of particle = ½ (2m) (v/2)2 = mv2 / 4

The energy is less than before since this is an inelastic collision.

If Δm = 4mv,

Final momentum of particle = +2mv – 4mv = – 2mv

The particle moves away from the wall. Mass of the particle = 2m, so the velocity of the

particle is now –v so that the momentum = (2m) (–v) = – 2mv

Final kinetic energy of particle = ½ (2m) (v)2 = mv2

Here, the energy is conserved, which is incorrect for an inelastic collision

Question 772: [Dynamics > Collisions]

Diagram shows a particle P, travelling at speed v, about to collide with a stationary particle Q

of the same mass. The collision is perfectly elastic.

Which statement describes motion of P and of Q immediately after the collision?

A P rebounds with speed ½ v and Q acquires speed ½ v.

B P rebounds with speed v and Q remains stationary.

C P and Q both travel in the same direction with speed ½ v.

D P comes to a standstill and Q acquires speed v.

Reference: Past Exam Paper – June 2011 Paper 12 Q13

Solution 772:

Answer: D.

For a perfectly elastic collision, both momentum and energy are conserved.

From the law of conservation of momentum,

Sum of momentum before collision = Sum of collision after collision

Before collision,

Sum of momentum, p = mv + 0 = mv

Total kinetic energy = ½ mv2

After collision,

Consider choice A,

Sum of momentum, p = m(0.5v) + m(-0.5v) = 0.

[A is incorrect]

Consider choice B,

Sum of momentum, p = - mv

[B is incorrect]

Consider choice C,

Sum of momentum, p = m(0.5v) + m(0.5v) = mv

Total kinetic energy = ½ m (0.5v)2 + ½ m (0.5v)2 = 0.25mv2

Consider choice D,

Sum of momentum, p = m(0) + mv = mv

Total kinetic energy = ½ mv2

Both momentum and energy are conserved

[C is incorrect]