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.

- 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.

A \(5\,\hat i\;m/sec\)

B \(3\,\hat i\;m/sec\)

C \(2\,\hat i\;m/sec\)

D \(6\,\hat i\;m/sec\)

- 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.

A \(3\,m/s\) towards left

B \(3\,m/s\) towards right

C \(5\,m/s\) towards right

D \(5\,m/s\) towards left

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

- 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.

- Draw two separate diagrams to show the system before and after the event.
- At maximum height, velocities are same.
- Choose the direction in which momentum can be conserved.
- Write down momentum conservation equation.
- Write down energy conservation equation.

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

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

C \(\dfrac{mv^2}{2g}\)

D \(\dfrac{Mv^2}{2g}\)

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

- 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.

- Draw two separate diagrams to show the system before and after the event.
- At maximum height, velocities are same.
- Choose the direction in which momentum can be conserved.
- Write down momentum conservation equation.
- Write down energy conservation equation.

A \(\sqrt {\dfrac{2 \; mgR}{M}}\)

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

C \(\sqrt{\dfrac{mgR}{m^2 + m}}\)

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

- 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.

- 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.

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

- 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.

A \(\sqrt {30}\,m\)

B \(\sqrt {20}\,m\)

C \(\sqrt {10}\,m\)

D \(\sqrt {40}\,m\)

- 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.

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.

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