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Work, Energy, and Power — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsWork, Energy, and Power5 Marks
Question
Define elastic and inelastic collisions. Write their basic characteristics. A bullet is fired into a block of wood. If it gets totally embedded in it and the system moves together as one entity, then state what happens to the initial kinetic energy and linear momentum of the bullet?
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Answer

Elastic collision: A collision in which there is absolutely no loss of kinetic energy is called elastic collision.

Characteristics :

  1. The linear momentum is conserved.
  2. Total energy of system is conserved.
  3. Kinetic energy is conserved.
  4. Forces involved during elastic collisions must be conservative forces.
Inelastic collision: A collision in which there occurs some loss of kinetic energy is called inelastic collision.

Characteristics :

  1. Linear momentum is conserved.
  2. Total energy is conserved.
  3. K.E. is not conserved.
  4. Some or all forces involved may be non-conservative.
Consider two particles A and B of masses m1 and m2 moving with initial velocities u1 and u2 along x-axis. They collide and move as one entity. Let V be the common velocity when they move as single mass.

m1 = Mass of bullet

u1 = Initial velocity of bullet

m2 = Mass of wood.

u2 = Initial velocity of wood = 0

According to conservation of momentum.

m1 u1 + m2 u2 = ( m1 + m2 )V

u2 = 0 

$\text{V}=\frac{\text{m}_1\text{u}_2}{\text{m}_1+\text{m}_2}\cdots{\text{i}}$

Initial K.E, $\text{k}_\text{i}\frac{1}{2}\text{m}_1\text{u}_1^2$

Final K.E., $\text{k}_\text{f}=\frac{1}{2}(\text{m}_1+\text{m}_2)\text{V}^2$

$\frac{\text{K}_\text{f}}{\text{K}_\text{i}}=\frac{\frac{1}{2}(\text{m}_1+\text{m}_2)\text{V}^2}{\frac{1}{2}\text{m}_1\text{u}_1^2}$

$=\Big(\frac{\text{m}_1+\text{m}_2}{\text{m}_1}\Big)\frac{\text{V}^2}{\text{u}_1^2}\ \cdots\text{(ii)}$

From (i) and (ii)

$\frac{\text{K}_\text{f}}{\text{K}_\text{i}}=\Big(\frac{\text{m}_1\text{m}_2}{\text{m}_1}\Big)\Big(\frac{\text{m}_1}{\text{m}_1+\text{m}_2}\Big)^2$

$=\frac{\text{m}_1}{\text{m}_1+\text{m}_2}$

When, $\text{K}_\text{f}<\text{K}_\text{i}$ there is loss in K.E.

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