Informative line

### Conservation Of Momentum And Mechanical Energy

Practice relative velocity problems with examples, learn kinetic energy when work done by external forces and non-conservative internal forces are zero, then total mechanical energy of the system remains constant.

# Relative Velocity

• We know that a quantity has its maximum or minimum value when its derivative is zero.

For ex- Velocity of a body is maximum or minimum when acceleration of that body is zero. $$(a=0)$$

• Similarly, relative displacement (separation) between two objects is maximum or minimum when its derivative i.e., relative velocity between them is zero.

Note : Relative velocity of two objects is said to be zero, if both of them have same speed and same direction.

• Now let us consider a situation, as shown in figure.
• Two blocks attached by an ideal spring, are kept on a smooth horizontal surface, as shown in figure.
• One of them is given a velocity $$'v'$$.

• In both cases, when extension is maximum or minimum, relative velocity of both blocks is zero.

Thus  $$\vec v_{AB}=0$$

$$\vec v_{AB}=\vec v_A-\vec v_B$$

$$0=\vec v_A-\vec v_B$$

$$\vec v_A=\vec v_B$$

$$\therefore$$  Velocity of both the blocks is same.

• Assume speed of $$A$$ is $$5\,m/s$$ towards right at the time of maximum compression then, speed of $$B$$ at that time will also be $$5\,m/s$$ towards right.

#### Two blocks attached by an ideal spring, are kept on a smooth horizontal surface, as shown in figure. Now $$A$$ is given a velocity of $$10\,m/s$$ towards right. If velocity of $$A$$ is $$3\,m/s$$ towards right at the time of maximum compression, then calculate the velocity of $$B$$ at that time.

A $$5\,\hat i\;m/sec$$

B $$3\,\hat i\;m/sec$$

C $$2\,\hat i\;m/sec$$

D $$6\,\hat i\;m/sec$$

×

At the time of maximum compression, i.e., minimum separation between blocks, their relative velocity is zero.

Thus,

Relative velocity $$\vec v_{AB}=0$$

$$\Rightarrow\;\vec v_A-\vec v_B=0$$

$$\Rightarrow\;\vec v_A=\vec v_B$$

Since velocity of $$A$$ is $$3\;m/sec$$ towards right, so velocity of $$B$$ is also $$3\;m/sec$$ towards right.

$$\vec v_B=3\,\hat i\,m/sec$$

### Two blocks attached by an ideal spring, are kept on a smooth horizontal surface, as shown in figure. Now $$A$$ is given a velocity of $$10\,m/s$$ towards right. If velocity of $$A$$ is $$3\,m/s$$ towards right at the time of maximum compression, then calculate the velocity of $$B$$ at that time.

A

$$5\,\hat i\;m/sec$$

.

B

$$3\,\hat i\;m/sec$$

C

$$2\,\hat i\;m/sec$$

D

$$6\,\hat i\;m/sec$$

Option B is Correct

# Relative Velocity of Block Placed on a Wedge

• Consider a system consisting of a block and a wedge.
• In the figure shown, a block $$A$$ with $$v$$, is pushed towards a wedge $$B$$.
• All the surfaces are smooth.

• Now, we observe the complete event from the frame of wedge $$B$$.
• From the frame of wedge $$B$$, we see that at maximum height, speed of block $$A$$ with respect to wedge $$B$$ is zero.
• That is block $$A$$ will be at maximum height, when its velocity relative to wedge $$B$$ is zero.
• For a observer on ground, when $$A$$ is at maximum height, the velocity of both $$A$$ and $$B$$ is same.

#### In the figure shown, a block $$A$$ is pushed towards a wedge $$B$$ with a velocity of $$5\,m/s$$, towards right. A wedge is moving with $$5\,m/sec$$ towards left. Finally when block $$A$$ is at maximum height on wedge $$B$$, then wedge $$B$$ will start moving with a speed of $$3\,m/sec$$ towards right. What is the velocity of block $$A$$ at the time when it is at maximum height?

A $$3\,m/s$$ towards left

B $$3\,m/s$$ towards right

C $$5\,m/s$$ towards right

D $$5\,m/s$$ towards left

×

Block $$A$$ will be at maximum height when its relative velocity with respect to wedge is zero.

Since, $$\vec v_{A/B}=0$$,

$$\vec v_B=3\,m/s$$ towards right

So, $$\vec v_{A/B}=\vec v_A-\vec v_B$$

$$\Rightarrow\;\vec v_A=\vec v_{A/B}+\vec v_B$$

$$\Rightarrow\;\vec v_A=0+3\,m/s$$

$$\Rightarrow\;\vec v_A=3\,m/s$$ towards right

### In the figure shown, a block $$A$$ is pushed towards a wedge $$B$$ with a velocity of $$5\,m/s$$, towards right. A wedge is moving with $$5\,m/sec$$ towards left. Finally when block $$A$$ is at maximum height on wedge $$B$$, then wedge $$B$$ will start moving with a speed of $$3\,m/sec$$ towards right. What is the velocity of block $$A$$ at the time when it is at maximum height?

A

$$3\,m/s$$ towards left

.

B

$$3\,m/s$$ towards right

C

$$5\,m/s$$ towards right

D

$$5\,m/s$$ towards left

Option B is Correct

# Calculation of Maximum Height when a Block is Projected towards a Wedge at Rest

### Condition for Conservation of Momentum

• When no external force acts on the system then the momentum of the system is conserved.

### Condition for Conservation of Mechanical Energy

• When there is no loss of energy in the system or none of the condition like jumping or landing, explosion jerk, kinetic friction and non-conservation force are present, then the mechanical energy of the system is conserved.
• Consider a situation in which a block A with velocity v is pushed towards a wedge, as shown in figure.
• All the contact surface are frictionless.

• To find out maximum height and velocity in above mentioned situation or similar situation, following steps to be followed.
1. Draw two separate diagrams to show the system before and after the event.
2. At maximum height, velocities are same.
3. Choose the direction in which momentum can be conserved.
4. Write down momentum conservation equation.
5. Write down energy conservation equation.

#### A block of mass m is projected with speed v towards a wedge B of mass M which is at rest, as shown in figure. If all the surfaces are smooth, then calculate the maximum height attained by the block on the wedge.

A $$\left(\dfrac{M}{M+m}\right)\dfrac{v^2}{2g}$$

B $$(M+m)\dfrac{v^2}{2g}$$

C $$\dfrac{mv^2}{2g}$$

D $$\dfrac{Mv^2}{2g}$$

×

Since at maximum height velocity is same, so

$$v_1^{'} = v_2^{'} = v^{'}$$

Since no external force is acting in horizontal direction, so momentum can be conserved in horizontal direction.

Applying momentum conservation equation in horizontal direction,

$$P_i = P_f$$

$$mv = (M+m)v^{'}$$

$$v^{'} = \dfrac{mv}{M+m}\;\;— (1)$$

There is no loss of energy, so applying energy conservation principle,

$$\dfrac{1}{2}mv^2 = \dfrac{1}{2} (M+m)(v^{'2}) + mgh$$

$$mv^2 = (M+m)(v^{'})^2 + 2 \; mgh$$

$$mv^2 = (M+m) × \left(\dfrac{mv}{M+m}\right)^2 + 2 \; mgh\;\;[from (1)]$$

$$mv^2 = \dfrac{m^2v^2}{M+m} + 2\; mgh$$

$$v^2 = \dfrac{mv^2}{M+m} + 2 \; gh$$

$$\Rightarrow 2 gh = v^2 \left[\dfrac{M+ m - m}{M+m}\right]$$

$$\Rightarrow h_{max} = \left[\dfrac{M}{M+m}\right] \dfrac{v^2}{2g}$$

### A block of mass m is projected with speed v towards a wedge B of mass M which is at rest, as shown in figure. If all the surfaces are smooth, then calculate the maximum height attained by the block on the wedge.

A

$$\left(\dfrac{M}{M+m}\right)\dfrac{v^2}{2g}$$

.

B

$$(M+m)\dfrac{v^2}{2g}$$

C

$$\dfrac{mv^2}{2g}$$

D

$$\dfrac{Mv^2}{2g}$$

Option A is Correct

# Conservation of Momentum and Mechanical Energy

### Condition for Conservation of Momentum

• When no external force acts on the system then the momentum of the system is conserved.

### Condition for Conservation of Mechanical Energy

• When there is no loss of energy in the system or none of the condition like jumping or landing, explosion, jerk, kinetic friction and non-conservation force are present, then the mechanical energy of the system is conserved.
• Consider a situation in which a block A with velocity v is pushed towards a wedge, as shown in figure.
• All the contact surface are frictionless.

• To find out maximum height and velocity in above mentioned situation or similar situation, following steps to be followed.
1. Draw two separate diagrams to show the system before and after the event.
2. At maximum height, velocities are same.
3. Choose the direction in which momentum can be conserved.
4. Write down momentum conservation equation.
5. Write down energy conservation equation.

#### A block A of mass m is kept inside a wedge B of mass M and radius R at position 1. If all the surfaces are smooth, then calculate the velocity of wedge when block is at position 2.

A $$\sqrt {\dfrac{2 \; mgR}{M}}$$

B $$\sqrt{\dfrac{2 \; mgR}{M^2 + m}}$$

C $$\sqrt{\dfrac{mgR}{m^2 + m}}$$

D $$\sqrt{\dfrac{2\,m^2 gR}{M^2 + Mm}}$$

×

Since there is no external force in horizontal direction, so momentum can be conserved in horizontal direction.

Applying conservation of momentum in x-direction,

$$P_i = P_f$$

$$0=mu\hat i - Mv\hat i$$

$$mu = mv$$

$$u = \dfrac{Mv}{m}$$

Since there is no loss of energy, so mechanical energy is conserved.

Applying energy conservation.

$$KE_i + PE_i = KE_f + PE_f$$

$$\Rightarrow 0 + 0 = \dfrac{1}{2} mu^2 + \dfrac{1}{2} Mv^2 - mgR$$

$$\Rightarrow 0 = \dfrac{1}{2} m\left(\dfrac{Mv}{m}\right)^2 + \dfrac{1}{2}\; Mv^2 - mgR$$

$$\Rightarrow mgR = \dfrac{1}{2} \; \dfrac{M^2}{m}v^2 + \dfrac{1}{2} Mv^2$$

$$\Rightarrow 2m^2gR = (M^2 + Mm)v^2$$

$$\Rightarrow v^2 = \dfrac{2m^2gR}{M^2 + Mm}$$

$$\Rightarrow v = \sqrt {\dfrac{2m ^2gR}{M^2 + M m}}$$

### A block A of mass m is kept inside a wedge B of mass M and radius R at position 1. If all the surfaces are smooth, then calculate the velocity of wedge when block is at position 2.

A

$$\sqrt {\dfrac{2 \; mgR}{M}}$$

.

B

$$\sqrt{\dfrac{2 \; mgR}{M^2 + m}}$$

C

$$\sqrt{\dfrac{mgR}{m^2 + m}}$$

D

$$\sqrt{\dfrac{2\,m^2 gR}{M^2 + Mm}}$$

Option D is Correct

# Conservation of Mechanical Energy

• When work is done only by conservative internal force, i.e., work done by external forces and non-conservative internal forces are zero, then total mechanical energy of the system remains constant.

$$\Rightarrow\;\text{Kinetic Energy (K.E.)}+\text{Potential Energy (U) = Constant}$$

Note :

• Kinetic energy $$(KE)$$ is maximum when potential energy $$(U)$$ is minimum.
• Kinetic energy $$(KE)$$ is minimum when potential energy $$(U)$$ is maximum.

#### A block is released from $$A$$, as shown in figure. If all the surfaces are smooth, then kinetic energy of the system is maximum when the block is at

A $$A$$

B $$B$$

C $$C$$

D

×

The system will have maximum kinetic energy when it has minimum potential energy.

Since the system has minimum potential energy at $$B$$, therefore the system will have maximum kinetic energy at $$B$$.

### A block is released from $$A$$, as shown in figure. If all the surfaces are smooth, then kinetic energy of the system is maximum when the block is at

A

$$A$$

.

B

$$B$$

C

$$C$$

D

Option B is Correct

# Common Velocity at the Time of Maximum Compression

• Consider two blocks A and B attached by an ideal spring having spring constant k, are kept on a smooth horizontal surface, as shown in figure.

• Now block A is given a velocity v, as shown.

• Since there is no net external force in horizontal direction therefore, two points has to be noted here,

a) Linear momentum in horizontal direction is conserved.

b) There is no loss of energy, so mechanical energy is conserved.

#### Two blocks A and B, of masses 5 kg and 10 kg respectively, attached by an ideal spring with spring constant $$k= 100\,N/m$$  are kept on a smooth horizontal surface, as shown in figure. If block A is given as velocity of $$10\,m/s$$, then calculate the speed of A and B at the time of maximum compression.

A $$\dfrac{10}{3} \hat i \,m/sec, \,\,\,\dfrac{10}{3} \hat i \,m/sec$$

B $$\dfrac{10}{3} \hat i \,m/sec, \,\,\,\dfrac{5}{3} \hat i \,m/sec$$

C $$\dfrac{8}{3} \hat i \,m/sec, \,\,\,\dfrac{5}{3} \hat i \,m/sec$$

D $$\dfrac{-10}{3} \hat i \,m/sec, \,\,\,\dfrac{10}{3} \hat i \,m/sec$$

×

At the time of maximum compression both the blocks will have same velocity.

Using the principle of conservation of momentum in x- direction,

Initial momentum = Final momentum

$$P_i = P_f$$

$$(5×10)\hat i +(10×0)\hat i = (5×u)\hat i +(10×u)\hat i$$

$$\Rightarrow 50 \,\hat i = 15 \,u \,\hat i$$

$$\Rightarrow u = \dfrac{50}{15} \,m/s$$

$$u = \dfrac{10}{3}\, m/s$$

So, at the maximum compression both of them will move with a speed $$u= \dfrac{10}{3} \,m/s$$ towards right.

### Two blocks A and B, of masses 5 kg and 10 kg respectively, attached by an ideal spring with spring constant $$k= 100\,N/m$$  are kept on a smooth horizontal surface, as shown in figure. If block A is given as velocity of $$10\,m/s$$, then calculate the speed of A and B at the time of maximum compression.

A

$$\dfrac{10}{3} \hat i \,m/sec, \,\,\,\dfrac{10}{3} \hat i \,m/sec$$

.

B

$$\dfrac{10}{3} \hat i \,m/sec, \,\,\,\dfrac{5}{3} \hat i \,m/sec$$

C

$$\dfrac{8}{3} \hat i \,m/sec, \,\,\,\dfrac{5}{3} \hat i \,m/sec$$

D

$$\dfrac{-10}{3} \hat i \,m/sec, \,\,\,\dfrac{10}{3} \hat i \,m/sec$$

Option A is Correct

# Steps to Calculate the Maximum Compression and Elongation in Spring

• Consider two blocks A and B attached by an ideal spring with spring constant k, are kept on a smooth horizontal surface, as shown in figure.

• Now block A is given a velocity v, as shown.

• Since, there is no net external force in horizontal direction therefore, two points need to be noted here,

a) Linear momentum in horizontal direction is conserved.

b) There is no loss of energy, so mechanical energy is conserved.

Consider two blocks A and B attached by an ideal spring are kept on a smooth horizontal surface, as shown in figure.

To calculate maximum compression or elongation in the spring following steps are to be followed:

1. Draw two separate diagram to show system before and after event.

2. Choose the direction in which momentum is conserved.

3. Apply momentum conservation equation.

4. Apply mechanical energy conservation equation.

#### Two blocks A (5 kg) and B (10 kg) attached by an ideal light spring (k = 100 N/m) are kept on a smooth horizontal surface. Calculate maximum elongation in the spring.

A $$\sqrt {30}\,m$$

B $$\sqrt {20}\,m$$

C $$\sqrt {10}\,m$$

D $$\sqrt {40}\,m$$

×

At the time of maximum elongation or compression velocity of both the blocks are same.

Since there is no net external force in horizontal direction, therefore we can conserve linear momentum in horizontal direction.

Applying momentum conservation equation

Initial momentum in horizontal direction = Final momentum in horizontal direction

$$20×10\hat i-10×5\hat i= (5+10)u\,\hat i$$

$$\Rightarrow150=15\,u$$

$$\Rightarrow u=10\,m/sec$$

Energy conservation equation

$$\dfrac {1}{2}×5×10^2+ \dfrac {1}{2}×10×(20)^2= \dfrac {1}{2}×k\,x^2+ \dfrac {1}{2}(5+10)×10^2$$

$$\Rightarrow 500+4000=100×x^2+1500$$

$$\Rightarrow 100\,x^2=4500-1500$$

$$\Rightarrow 10\,x^2=450-150$$

$$\Rightarrow x^2=\dfrac {300}{10}$$

$$\Rightarrow x=\sqrt {30}\,m$$

### Two blocks A (5 kg) and B (10 kg) attached by an ideal light spring (k = 100 N/m) are kept on a smooth horizontal surface. Calculate maximum elongation in the spring.

A

$$\sqrt {30}\,m$$

.

B

$$\sqrt {20}\,m$$

C

$$\sqrt {10}\,m$$

D

$$\sqrt {40}\,m$$

Option A is Correct

# Spring with Two Blocks System

• Consider two blocks A and B attached by an ideal spring with spring constant k, are kept on a smooth horizontal surface, as shown in figure.

• Now block A is given a velocity v, as shown.

• Since, there is no net external force in horizontal direction therefore, two points need to be noted here,

a) Linear momentum in horizontal direction is conserved.

b) There is no loss of energy, so mechanical energy is conserved.

Consider two blocks A and B attached by an ideal spring are kept on a smooth horizontal surface, as shown in figure.

To calculate maximum compression or elongation in the spring following steps are to be followed:

1. Draw two separate diagram to show system before and after event.

2. Choose the direction in which momentum is conserved.

3. Apply momentum conservation equation.

4. Apply energy conservation equation.

#### Two blocks A and B, of masses 5 kg and 10 kg respectively, attached by an ideal spring constant k = 100 N/m, are kept on a smooth horizontal surface. If 5 kg block is given a velocity of 10 m/s in rightward direction, as shown in figure, calculate maximum velocity of 10 kg block.

A $$1 \; m/s \; or \; \dfrac{10}{3} \; m/s$$

B $$20 \; m/s \; or \; 10 \; m/s$$

C $$0 \; m/s \; or \; \dfrac{20}{3} \; m/s$$

D $$\dfrac{10}{3 }\; m/s \; or \; \dfrac{8}{3} \; m/s$$

×

The 10 kg block will have maximum velocity when spring will be relaxed.

Let, at the time of maximum velocity of 10 kg block x=extension in the string.

Since there is no net external force in horizontal direction, therefore we can conserve momentum in horizontal direction.

Momentum conservation equation,

Initial momentum in horizontal direction = Final momentum in horizontal direction

$$10 × 5 \; \hat i = 5 × u \; \hat i + 10 × v \; \hat i$$

$$\Rightarrow 5 0 = 5 \; u + 10 \; v$$

$$\Rightarrow10 = u + 2 v \; — (1)$$

Energy conservation,

$$\dfrac{1}{2} × 5 × (10)^2 = \dfrac{1}{2} × 5 × u^2 + \dfrac{1}{2} × 10 × v^2$$

$$\Rightarrow 500 = 5× (10-2v)^2 + 10 v^2$$

$$\Rightarrow100 = 100 + 4 v^2 - 40 v + 2 v^2$$

$$\Rightarrow6 v^2 = 40 v$$

$$\Rightarrow v = 0 \;or\; v = \dfrac{20}{3}\; m/s$$

### Two blocks A and B, of masses 5 kg and 10 kg respectively, attached by an ideal spring constant k = 100 N/m, are kept on a smooth horizontal surface. If 5 kg block is given a velocity of 10 m/s in rightward direction, as shown in figure, calculate maximum velocity of 10 kg block.

A

$$1 \; m/s \; or \; \dfrac{10}{3} \; m/s$$

.

B

$$20 \; m/s \; or \; 10 \; m/s$$

C

$$0 \; m/s \; or \; \dfrac{20}{3} \; m/s$$

D

$$\dfrac{10}{3 }\; m/s \; or \; \dfrac{8}{3} \; m/s$$

Option C is Correct