Rankers Physics

Principal of Conservation of Energy: Practice Problem & Solution

The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is Principal of Conservation of Energy diagram: The block of mass M moving on the
Zero
ML²/K
(MK)½ L
KL²/2M

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Principal of Conservation of Energy. Understanding the underlying formula is key to arriving at the correct answer below:

From Principal of conservation of Energy in absence of non-conservative forces,

Ui+ Ki= Uf+ Kf

½KL²+ 0 = ½mv²

⇒v= √(KL²/m)

Momentum is P= m.v = m. √(KL²/m) =(mK)½ L

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. A system absorbs 600 J of the system works equivalent to –900 J by The value of ΔE for the system is :
2. 250 N force is required to raise 75 kg mass from a pulley. If rope is pulled 12 m then the load is lifted to 3 m, the efficiency of pulley system will be: (2001)
3. A child on a swing is 1 m above the ground at the lowest point and 6 m above the ground at the highest point. The horizontal speed of the child at the lowest point of the swing is approximately
4. A mass m slips along the wall of a hemispherical surface of radius R. The velocity at the bottom of the surface is
5. A child is sitting on a swing. Its minimum and maximum heights from the ground is \(0.75\text{ m}\) and \(2\text{ m}\) respectively, its maximum speed will be: (2001)