A block of mass 200 g is suspended through a vertical spring. The spring is stretched by 1 ·0

A block of mass 200 g is suspended through a vertical...

Key Concepts

Mass–Spring Oscillations

Mass–spring oscillations involve a mass attached to a spring undergoing periodic motion about an equilibrium position. By analyzing the energy transformation between kinetic, gravitational, and spring potential energies during the motion, one can predict extreme positions such as the maximum extension or compression of the spring.

Conservation of Mechanical Energy in Oscillatory Systems

After the collision, the system (block plus attached particle) undergoes oscillatory motion. In the absence of non-conservative forces, the mechanical energy of the system is conserved. The kinetic energy of the system is converted into gravitational potential energy (due to further descent) and elastic potential energy stored in the spring, which allows one to determine the maximum extension of the spring.

Hooke's Law and Equilibrium

Hooke's Law states that the force exerted by a spring is directly proportional to its extension or compression from its natural length, expressed as F = kx. In an equilibrium situation where the spring holds a mass, the gravitational force (weight) is balanced by the spring force, allowing one to determine the spring constant k by equating mg with kx.

Momentum Conservation in Inelastic Collisions

In an inelastic collision, where particles stick together upon impact, kinetic energy is not conserved but momentum is. The combined mass immediately after the collision moves with a velocity determined by the conservation of linear momentum. This principle is crucial for analyzing the dynamics following the impact before energy transformations in the oscillatory system take place.

Transcript

00:01 In this situation given data is that is there is a figure given so first i will draw this figure so in the figure it is saying that a block of mass 200 kg sorry 200 gram is suspended through a vertical spring and this spring is due to 200 gram is stretched by 1 centimeter that means it will stretch by 1 centimeter you can see that here i had written the value that is after stretching this spring is going downward due to this block so consider that at this position the block is now present so now in the question it is saying that there is a particle of mass 120 which is dropped from block at a height of 45 cm so consider that this is a particle whose might is sorry mass is given that is one 20 gram and height will be from block is that is 45 cm so now in the kitchen it is saying that this particle is going to drop the vertically so after it it will stick with this block that means after dropping the new condition will be that is this particle is going to stick with this block and both are going to with same velocity in a downward direction and so due to going downward the spring is going i can see that now spring is extending so in this question it is asking that we have to find the maximum extension of spring after going collision.

02:28 So this is the question.

02:31 I had explained every data by making a figure.

02:35 So now let's start answering this question.

02:40 So i can say that in the answer the mass of block is given that is capital m is equal to 200 gram.

02:55 Or i can say that in kg it will be written as it is 0 .20 kg and mass of particle is given that is small m is equal to 120 gram or i can say that in kg it will be written as that is 0 .12 kg so and height of the particle from first block is given that is h is equal to 45 centimeter or i can say that 0 .45 meter so now according to cushion as as a block attains equilibrium the spring is stretched by one centimeter you can see that here the block is going downward due to weights and due to its own weight and after it will take equilibrium position so due to taking equilibrium position the spring is stretched by one centimeter so i can say that the next step will be written as that is own weight will be equal to spring force.

04:22 Means the weight of block will be equal to spring force.

04:28 So put the well data, so it will written as mg will be that is m is mass of block that will be 0 .2 kg multiplied by g is that is gravity and we know that the value is 10 meter per second square.

04:44 So here i can say that 0 .2 multiplied by 10 is equal to k multiplied by x x x is stretching how much the spring is stretched by going downward due to for given block this block so in the question it is saying at one centimeter so here i will write that is x is equal to one centimeter or i can see that in the meter it will be written as it is 0 .01 so here substitute the value of x that is 0 .01 now simply it then i will get the spring constant k that is equal to 200 newton per meter.

05:35 So now next step is i can say that you can the particle is dropping downward vertically so the speed of the particle will be that is or i can say that velocity with the particle which is striking with block is given by the formula that is, sorry, you u is equal to sky root of 2g h.

06:19 And i can say that it will be written as it is u is equal to sky root of 2 multiplied by g will be that is 10 multiplied by h will be that is 0 .45 meter.

06:33 So i can say that put the value of h here that is 0 .45 meter...

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