A bullet is fired and gets embedded in a block kept on table. If table is frictionless, then
A ball with a velocity of 2 m/s collides head-on with a stationary ball of double mass. Using the coefficient of restitution of 0.5, the final velocities after the collision are 0 m/s for the first ball and 1 m/s for the second ball.
Chapter 5 of Class 11 Physics covers Work Energy and Power. It explains the concepts of work done by a force energy transformations and the principle of conservation of energy. The chapter also discusses kinetic energy potential energy and power along with their formulas and units. Real-life applications and numerical problems enhance understanding of these fundamental concepts for the CBSE exam.
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In the scenario where a bullet is fired and gets embedded in a block on a frictionless table, we assess the conservation laws that apply to the situation.
1. Momentum Conservation:
– During the collision, the system or the bullet along with the block is isolated; no external force acts on the system in the horizontal direction (since the table is frictionless). Hence, momentum is conserved.
– Before the collision, the block is at rest, and the bullet is moving with its momentum. After the bullet gets embedded in the block, the total momentum is conserved before the collision to be equal to the total momentum after the collision.
2. Kinetic Energy Conservation:
– In this inelastic collision, kinetic energy is not conserved. Some of the kinetic energy from the bullet goes into internal energy during the collision (such as heat, sound, and deformation of the bullet and block).
3. Potential Energy Conservation:
– Potential energy is not involved in this problem since it does not change due to the fact that the bullet is embedded into the block horizontally on a frictionless table.
Final Answer:
In this case, momentum is conserved.
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