Conservation of linear momentum and energy experiment by...

Conservation of Linear Momentum and Energy Experiment by using PHET Simulation Please help me finish the rest of the table: Part 2: Inelastic Collision: Use the same link from PHET Colorado Simulation. https://phet.colorado.edu/sims/html/collision-lab/latest/collision-lab_en.html CHANGE: Elasticity to 0% by dragging the blue triangle to the left Select: m1 = 3 kg and m2 = 2 kg KEEP: *position 1 (0.5 m) and *position 2 (1.5 m) Select: *v1i = 2.00 m/s and *v2i = 0 Click on Play, then after collision Pause Record the value of vf into tables 4 & 5. Calculate the values of Pi, Pf, Pf/Pi, Ki, Kf and Kf/Ki and record them into tables 4 & 5. Repeat the experiment for three other different values of v1i and KEEP v2i = 0 as shown in tables 4 & 5 and record the values into tables 4 & 5. Table 4 V1i (m/s) | V2i (m/s) | Vf (m/s) | Pi = m1v1i (kg·m/s) | Pf = (m1 + m2)vf (kg·m/s) | Pf/Pi 2.00 | 0 | 1.2 | 6 | 6 | 1 2.50 | 0 | | | | 3.00 | 0 | | | | Table 5 V1i (m/s) | V2i (m/s) | Vf (m/s) | Ki = (0.5m1v1i^2) (J) | Kf = 0.5(m1 + m2)vf^2 (J) | Kf/Ki 2.00 | 0 | 1.2 | 6 | 3.6 | 0.6 2.50 | 0 | | | | 3.00 | 0 | | | |

Transcript

00:01 Hello everyone so let's start with the question.

00:01 So in a question we have a two mass that is 3kg and 2kg so suppose this is the mass m1 which is moving with the velocity of 2 meter per second and there is another mass that is m2 which is at rest and it's having a mass 2k g okay so as you can see m1 is greater than m2 so when this mass m1 is collide with this mass m2 so this is before the collision so after the collision so what what did happen that m2 will be stuck with m1 and they both the mass will move together in this direction with having a velocity vf okay so it so value will be what that is 1 .2 meter per second that is given in a question okay so this is the after collision and collision is a elastic collision elastic collision so here we have to find the value of final velocity for the table 4 and 5.

01:05 So this is table 4 and this is table 5.

01:13 So we have to find the velocity for this both the table.

01:16 Also we have to find initial and final momentum, the ratio of both final and initial momentum and kinetic energy of initial and final and the ratio of kinetic energy for final and initial condition.

01:28 So as you can see that for the table 4.

01:33 So we have to find the velocity for this table 4.

01:36 So what it will be? so by using the conservation of momentum, using conservation of momentum, we know the formula.

01:48 That is m1 v1 plus m2 u2.

01:53 So this is u1.

01:56 U1 equal to m2 v2 plus m1 v1 v1 plus m2 v2.

02:10 So here we have v1 initial v2 initial and vf.

02:15 So as the in before the collision it was at rest so this will be cancelled out so it will be 0.

02:23 So it will be m1.

02:24 M1 is what that is 2 3 kg and v1 that is given in a question that is 2 meter per second.

02:29 So it will be 2 equal to okay so in a question for this condition like for the initial condition for the initial velocity it is given in a question that 2 meter per second so after solving means after putting the value in this equation that is conservation momentum we get the final velocity that is 1 .2 meter per second as you can see so it will be 1 .2 meter per second and the initial momentum that is pi is what m1 v1 okay so if if i multiply this both the term okay so we get our answer that is that is the mass is 3 and initial as it is 2 so it will be 6 similarly for the final momentum it have a 2 value and its final velocity is what that is means for the final momentum so it will be what m1 plus m2 multiply by vf that is final velocity okay so we use this formula and by putting the value we get our answer that is 6 and its ratio will be 1 that is 6 divided by 6 okay so for the next that is the 2 .5 means initial condition means initial value for the velocity v1 that is 2 .5 equal to now this this will be zero okay that is given in a table now m1 will be so as you can see that after the collision the well means both the mass will be go together so it will be having a same velocity okay so here if i comment out the velocity that is here after the collision it will be what that is v equal to m1 plus m2 so after the collision the velocity become v1 equal to 2 so therefore 3 multiply by 2 .5 equal to v that we have to find that is vf and 3 plus 2 okay so after putting the value in the equation we get our final velocity that is 1 .5 meter per second so okay so put in the table in table 4 that is 1 .5 meter per second initial momentum same m1 v1 so m1 is what that is 3 kg and v1 is what 2 .5 meter per second so put after putting it we get our answer that is 7 .5 and pf that is final momentum so it is m1 plus m2 multiply by vf so m1 plus m2 is what 3 plus 2 5 kg multiply by vf okay multiply by vf that is 7 .5 after putting it we get our answer that is 7 .5 and its ratio will be 1 same for this means 3 meter per second so after putting it we get our final velocity that is 1 .5 8 and initial momentum will be 9 and the final momentum will be 9 and its ratio will be 1.

05:14 So this is the answer for the table 4...

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