Head On Collision Points: 20 A mass M2=5.90 kg slides on a...

Head On Collision Points: 20 A mass $M_2$=5.90 kg slides on a surface with negligible friction and collides head-on with mass $M_1$. $v_0$ $M_2$ $M_1$ $v_2$ $M_2$ $M_1$ After Before Before the collision, $M_2$ has a speed $v_0$=5.50 m/s and $M_1$ is at rest. After the collision, $M_2$ has speed $v_2$=2.90 m/s and $M_1$ has speed $v_1$=3.50 m/s, both in the initial direction of $M_2$ (see figure). Calculate the mass of $M_1$. Submit Answer Tries 0/100 For the collision above, select the one best answer for the statement below. The collision is inelastic. Submit Answer Tries 0/100

Transcript

00:01 Hi there, so for this problem, let's call the first one problem one.

00:05 We are told that an object with a mass that we're going to call lowercase m is equal to 500 grams.

00:11 So that is the same as 0 .5 kilograms.

00:17 And with an initial speed, so we're given its initial speed that we're going to call just simply as you.

00:30 And that is equal to 0 .2 meters per second.

00:39 And collides with another object with a mass that we're going to call them as capital m that is equal to 1 .5 kilograms.

00:50 And which was at rest before the collision, so its initial speed, we're going to call this initial speed prime, is equal to zero because we are told that it is at rest initially.

01:06 Now we need to calculate the resultant speed for an inelastic collision when they stick together.

01:13 So we need to obtain that speed that we're going to call capital b.

01:17 Now, since we know that this is, well, we need to apply conservation of momentum.

01:24 So the initial momentum is just simply the mass lowercase m times its initial speed, and then this is equal to the final condition where they stick together, so that will be the sum of the mass capital m and the mass lowercase m times the final speed.

01:42 So if we solve for that final speed, we will have that that is the mass times initial speed divided by the sum of the masses in here.

01:50 And now we just need to simply substitute the values in here.

01:55 So that will be 1 .5 kilograms plus 0 .5 kilograms and in the numerator we have 0 .5 kilograms times the initial speed that is equal to 0 .2 meters per second so from this we obtain a value of so the speed of the system after the collision is 0 .0 5 meters per second so that's a solution for part a of this problem.

02:34 Well, for the first problem in here...

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