In the picture below two blocks of mass m1 75 kg and m2 53...

In the picture below, two blocks of mass m₁ = 7.5 kg and m₂ = 5.3 kg will collide elastically. Block 1 starts with an initial speed of 2.5 m/s at a height of yᵢ = 1.4 m. Block 2 is initially stationary. The track is frictionless. a) What is the velocity of m₁ right before it collides with m₂? b) What is the velocity of m₁ right after it collides with m₂? c) What is the velocity of m₂ after the collision? 1. In the picture below, two blocks of mass m = 7.5 kg and m = 5.3 kg will collide elastically. Block 1 starts with an initial speed of 2.5 m/s at a height of yᵢ = 1.4 m. Block 2 is initially stationary. The track is frictionless. Momentum = mv a) What is the velocity of m right before it collides with m₂? b) What is the velocity of m right after it collides with m? c) What is the velocity of m after the collision?

Transcript

00:01 Right, this is a classic case of conservation of momentum.

00:03 Three blocks are on horizontal frictionless surface and block m1 is 4 kilograms that moves at a velocity of 2 .5 meter per second to the right.

00:13 And a block of mass 5 to kilograms moves over on the left with 2 .8 meter per second.

00:19 Two blocks collide and bounce off to each other, bounds off of each other.

00:24 And the block m2 moves over on the right with 2 .1 meter per second.

00:29 And m2 then collides with m3 of 2 kilogram mass and which was initially at rest and the two blocks stick together.

00:37 So it's a velocity of m1 after the collision, after its collision with m2 with both magnitude and directions.

00:42 So differently, like i said, so there is no external force.

00:46 So the momentum is being conserved.

00:53 If the momentum is being conserved, then the initial momentum, the initial situation is like this.

01:01 So this is the block m1, which is of 4.

01:04 Kilograms as moving over on the right with 2 .5 meter per second and the block m2 this is 5 kilogram which is moving over on the left with 2 .8 meter per second after the collision the block m1 where is that moving we do not know so let's consider it as we of perhaps let's call it a we of a and the other block is already moving on the right with 2 .1 meter per second and this is 5 kilograms okay so the initial momentum is going to be 4 times 2 .5 minus 2 .8 times 5 why minus because this is over on the other side and that's why the momentum is negative this comes out as 4 times 2 .5 minus 2 .8 times 5 which is coming as negative 4 and the final momentum is going to be 4 times v a plus 5 times 2 .1 so this is coming as this is coming as 10 .5 plus 4 times va so 10 .5 plus 4 times v sub a is equal to minus 4 because the momentum is conserved, subtracting both sides by 10 .5.

02:44 So we have this value added to 10 .5, which is negative, subtracting 10 .5 both sides, so that's negative 14 .5.

02:55 And dividing both sides by four, we have the value of v .a as 3 .625 meter per second with a negative sign.

03:06 So definitely the mass m1 is moving with 3 .625.

03:09 6 to 5 meter per second over on the left after the collision okay so what is a change in the kinetic energy of block m1 plus m2 just before and after the collisions that's also easier to find because it's just a matter of calculation so the initial kinetic energy of the system is half times four times 2 .5 square because the kinetic energy is half mv square plus the kinetic energy of this one so that's half times five times 2 .8 square which is coming as 2 .5 square times 2 plus 2 .5 times 2 .8 square.

03:50 I'm just doing with calculator.

03:52 So that is coming as 32 .1 meter, sorry, jules.

03:59 Final kinetic energy is going to be half times 4 times va.

04:04 What is va? 3 .625 .25 whole square plus half times 5 times 2 .1 square.

04:15 So that's going to come out as 3 .625 square times 2 plus 2 .5 times 2 .1 square.

04:27 So this is coming as 37 .31 jules...