Two blocks are free to slide along the frictionless wooden...

Two blocks are free to slide along the frictionless wooden track $A B C$ shown in Figure P9.19. The block of mass $m_1=$ 5.00 kg is released from A . Protruding from its front end is the north pole of a strong magnet, which is repelling the north pole of an identical magnet embedded in the back end of the block of mass $m_2=10.0 \mathrm{~kg}$, initially at rest. The two blocks never touch. Calculate the maximum height to which $m_1$ rises after the elastic collision.

Key Concepts

Conservation of Kinetic Energy in Elastic Collisions

An elastic collision is defined as a collision in which the total kinetic energy of the system is conserved. This characteristic, along with momentum conservation, provides a set of equations that can be solved simultaneously to determine the final velocities of objects after collision. Understanding this principle is essential for analyzing interactions where energy loss to deformation, sound, or heat is negligible.

Energy Conversion to Gravitational Potential Energy

After the collision, when an object moves upward along a frictionless incline or vertical track, its kinetic energy is converted into gravitational potential energy. This conversion is governed by the relation between kinetic energy and the potential energy stored due to its height in a gravitational field. Analyzing this energy transformation is key to determining the maximum height achieved by the object as it ascends against the force of gravity.

Conservation of Momentum

In collision problems, especially in isolated systems, the total momentum of the system remains constant before and after the collision. This concept allows us to relate the velocities of the objects before and after the collision, regardless of the forces acting during the very short time of impact. It is crucial in determining the post-collision speeds of the interacting masses in any type of collision.

Transcript

00:03 In this question, we are dealing with wooden track that is frictionless.

00:08 And then there are two blocks, m1 and m2.

00:16 Okay, and this is 5 meters.

00:24 So yeah, so what happens is that m1 is going to be released from ras and then as it moves down, it would, interact with m2 through magnetic interaction.

00:42 They have magnets put at the ends that they are going to be touching, not really touching, but almost touching.

00:52 So they are going to repel later.

00:55 And in this question, we want to find a maximum height to which m1 rises after the elastic collision.

01:02 Okay, so one thing we know is that, as m1 reaches the bottom is going to interact with m2 okay so and m1 is going to move with some speed we while m2 is at rest okay so to find the speed of block 1 to find the speed of m1 just before collision okay this is equal to square root 2 gh 2 times 9 .8 times 5 meters and calculate this you get 9 .90 meters per second.

01:55 So this is obtained through conservation of energy of the m1 earth system.

02:20 Then we have conservation of momentum for the two block system.

02:29 So we have m1b equals to m1b1f plus m2 v2f.

02:46 So here i'm going to take right to be positive.

02:55 So we have one equation from conservation of momentum.

02:59 We need another equation, which is from elastic collision.

03:04 We know that relative speed of approach is equal to relative speed of separation, which means that v1i vector minus v2i vector equals to v2f vector vector.

03:38 V2f vector.

03:39 Okay, so i'm going to take right to be positive again.

03:44 So we have v minus 0 equals to v2f minus v1f...

Please give Ace some feedback