A block of mass 2 ·0 kg is moving on a frictionless horizontal surface with a velocity of 1 ·0

A block of mass 2 ·0 kg is moving on a frictionless...

Key Concepts

Conservation of Momentum

In a closed system where external forces are negligible, the total momentum before an event is equal to the total momentum after the event. This principle is especially important when analyzing collisions or interactions between objects, such as when one moving object interacts with another stationary object, ensuring that the system’s overall momentum remains constant.

Conservation of Energy

In the absence of non-conservative forces like friction, the total mechanical energy of a system is conserved. During interactions involving springs and colliding masses, kinetic energy can be converted into potential energy (and vice versa). This principle enables one to equate the kinetic energy lost by the moving mass to the energy stored in the spring at the point of maximum compression.

Hooke's Law and Spring Potential Energy

Hooke's Law describes the force exerted by a spring as proportional to its displacement from equilibrium, with the spring constant being the proportionality factor. The potential energy stored in a compressed or stretched spring is given by ½ k x², where k is the spring constant and x is the displacement. This concept is critical for relating the energy transferred from the moving mass to the spring when it is compressed to its maximum extent.

Transcript

00:01 In this question given data is that is there is a figure given so first i will draw this figure so in this figure there is a hard surface given and on this hard surface there are two block is given so first block having mass that is given in the cushion that is 2 kg and similarly second block is having same mass that is 2 kg and there is a spring is attached to a second block and now in the kitchen it is saying that initially they both are rest but after some time this first block is going in this direction with the velocity that is given one meter per second so in the question it is asking that we have to find the maximum compression of spring and in the question the spring constant is given that is equal to 100 newton per meter so this data is given now let's start answering the solution so in the answer i will say that the mass of block is first for first block i can say that mass will be written as that is ma is equal to 2 kg and for mass of second block that is mb will be written as that is 2 kg.

01:43 Now initial velocity for first block will be written as that is va is equal to that is 1 meter per second and initial velocity for second block will be written as that is 0 because the second block is i consider that is vb is equal to 0 because the second block is a second block is you can see that it is at rest position so here i had written the value that is equal to zero now in the question spring constant is given that is k is equal to 100 newton per meter so now next step is as you see in the figure this first block is trying to move with one meter per second in this direction and taking a collision with second block which is initially at rest position so after collision the first block the first block the velocity is going to decrease continuously and at instant of time the wall system that is first block and this is spring and this second block moves together with common velocity that is i can see that that is considered capital v so i can say that here i will write that is after the collision wall system that is block first block plus spring plus second block moves together with a common velocity which i say that capital v here i had assumed this so now next step is here i'm using the law of energy conservation of energy so apply the law of conservation of energy that will say that total kinetic energy before collision will be equal to total kinetic energy after collision plus spring energy so the left side will be written as that is total kinetic energy before collision that is half m a plus sorry m a v a square plus half mb b b b square is equal to half m m .a.

05:08 V.

05:09 Square plus half m .a.

05:11 V.

05:12 Square sorry mv .v square plus spring energy that is half kx square.

05:26 So now simplify it.

05:32 So here x is a compression.

05:37 You can understand that here x is a compression of spring when the first block is trying to collide with second block...

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