HC Verma Solutions for Class 11 Chapter 9 focus on concepts like Centre of Mass, Linear Momentum, and Collisions. It explains conservation laws, impulse, types of collisions, and rocket propulsion. Detailed examples and problem-solving techniques ensure students grasp essential physics concepts effectively, building a strong foundation for advanced topics.
R = (∑miri) / (∑mi)p = mv.∑pinitial = ∑pfinale = (Relative velocity after collision) / (Relative velocity before collision)J = ∆p = F∆tv = ve ln(m0/m)Problem: Two particles of masses 3 kg and 5 kg are located at (2, 0) m and (0, 4) m, respectively. Find the centre of mass.
Solution:
xcm = (3 * 2 + 5 * 0) / (3 + 5) = 0.75 mycm = (3 * 0 + 5 * 4) / (3 + 5) = 2.5 m
Centre of Mass: (0.75, 2.5) m
Problem: A ball of mass 2 kg moving at 5 m/s collides elastically with another ball of mass 3 kg at rest. Find their velocities after the collision.
Solution:
Using conservation of momentum:
2 * 5 + 3 * 0 = 2v1 + 3v210 = 2v1 + 3v2 ...(1)
Using conservation of kinetic energy:
25 = v12 + (3/2)v22 ...(2)
Solve equations (1) and (2):
v1 = -1 m/s, v2 = 4 m/s
When two objects moving in opposing directions with the same momentum (p=mv) collide in a completely inelastic collision, they cling together. The momentum will be 0 because it is conserved. The kinetic energy (of motion) is largely transferred to heat because they are not moving after the contact.
In a two-particle system, the centre of mass divides the internal line connecting the two particles in the inverse ratio of their masses. If we know the centres of mass of the system’s parts and their masses, we can find the combined centre of mass in a system with a larger number of particles.
A simple explanation of the centre of mass with a diagram is provided in the section “Centre of Mass.” The centre of mass in a two-particle system divides the line connecting the two particles internally in the inverse ratio of their masses.