A ball of mass 0.200 kg has a velocity of 150 𝐢̂ m / s ; a ball of mass 0.300 kg has a velocity

A ball of mass 0.200 kg has a velocity of 150 𝐢̂ m / s ; a...

A ball of mass $0.200 \mathrm{kg}$ has a velocity of $150 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s} ;$ a ball of mass $0.300 \mathrm{kg}$ has a velocity of $-0.400 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s} .$ They meet in a head-on elastic collision. (a) Find their velocities after the collision. (b) Find the velocity of their center of mass before and after the collision.

A ball of mass $0.200 \mathrm{kg}$ has a velocity of $150 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s} ;$ a ball of mass $0.300 \mathrm{kg}$ has a velocity of $-0.400 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s} .$ They meet in
a head-on elastic collision. (a) Find their velocities after the collision. (b) Find the velocity of their center of mass before and after the collision.

Key Concepts

Conservation of Momentum

This principle states that in the absence of external forces, the total momentum of a system remains constant before and after a collision. It is fundamental in collision problems, allowing you to equate the total momentum of the bodies prior to impact to the total momentum after impact.

Conservation of Kinetic Energy

In elastic collisions, not only is momentum conserved, but the total kinetic energy of the system is also preserved. This additional constraint is essential for uniquely determining the final velocities in an elastic collision.

Elastic Collision

An elastic collision is one in which both momentum and kinetic energy are conserved. This type of collision often allows the use of specific formulas or relationships, such as the relative speed of approach equaling the relative speed of separation, to determine the final velocities of the colliding objects.

Center of Mass

The center of mass of a system represents the weighted average position of all the mass in that system, and its velocity is given by the momentum-weighted average of the individual velocities. In isolated systems, the velocity of the center of mass remains constant throughout interactions such as collisions.

Transcript

00:02 So one mass of 0 .2k is having velocity 150 icap meter per second and other mass of 0 .5 kg is having velocity of minus 0 .4 i cap meter per second okay and let us say after collision they both are moving in right direction this is what we are considering here its mass is 0 .2 kg and its mass is 0 .3 kg if you see clearly its velocity is minus 0 .4 iq it means it is moving in negative x direction with 0 .4 meter per second okay so in order to find these final velocities we will apply conservation of linear momentum so conservation of linear momentum initial momentum is equal to final momentum that is momentum before collision is equal to momentum after collision so simply we can write here mass into velocity that is 0 .2 into 150 minus 0 .3 into 0 .4 is equal to 0 .2 v1 plus 0 .3 v2 so on solving this expression we are getting 29 .88 is equal to 0 .2v1 plus 0 .3 v2 or other words we can write 2 v1 plus 3 v2 or other words we can write 2 v1 plus 3 v2 is equal to 298 let us say this is equation 1 so we know since the collision is head -on elastic collision so for head -on elastic collision coefficient of destitution value is 1 coefficient of restitution the value of e is 1 and e how we are finding the value of e it is the velocity of separation after collision over velocity of approach before collision so velocity of separation after collision is v2 minus v1 and velocity of approach before collision is 150 plus 0 .4 is equal to 1 so from here we are getting v2 minus v1 is equal to 150 .4 so if we multiply it by 2 it will become minus 2 1 plus 2 v2 is equal to 300 .2 let us say this is equation number two okay so if we add these two equations adding equation one and equation two what will happen we will get the value 5 v2 is equal to 599 .6 and from here we are getting the value of v2 as 119 .9 meter per second so if we put this value in equation number 2 then we can write v1 is equal to v2 minus 150 .4 from this equation okay so now we substitute the value of v2 here 119 .9 minus 150 .4 so that value of of v1 comes out to be minus 30 .5 meter per second so simply we can write the answer of first part as v1 is equal to minus 30 .5 i cap meter per second and v2 is equal to 119...

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