As shown in the figure, a metal ball with mass m2 is...

As shown in the figure, a metal ball with mass $m_2$ is initially at rest on a horizontal, frictionless table. A second metal ball with mass $m_1$ moving with a speed 2.00 m/s, collides with $m_2$. Assume $m_1$ moves initially along the +x-axis. After the collision, $m_1$ moves with speed 1.00 m/s at an angle of $\theta = 53.0^\circ$ to the positive x-axis. (Assume $m_1 = 0.200$ kg and $m_2 = 0.300$ kg.) (a) Determine the speed (in m/s) of the 0.300 kg ball after the collision. m/s (b) Find the fraction of kinetic energy transferred away or transformed to other forms of energy in the collision. $\frac{|\Delta K|}{K_i} = $

Transcript

00:01 Hi guys, now in this film.

00:01 We have an object first which is m1 and this is m2.

00:06 The m1 is moving at a speed of what this is 3 meter per second, sorry 2 meter per second and this m2 is at the rest.

00:17 This is u is equal to 0 .m 1 has a mass of a 0 .2 kg this m2 has a mass of what 0 .3 kg.

00:31 Now collision is happening here let me say that this is collision happening at this point.

00:37 This is a collision point collision point after collision what will be happening these two blocks are moving like this if i'm saying that this is my horizontal line i have and here i have a horizontal line that i have now this is moving at an angle of what this at the angle of let me draw this structure this not lose it's not look so nice this is the horizontal line that i have and with respect to this line let me draw with a dotted line line first one is moving like this second one is moving like this these are the objects that i have here i have and like this this is at the angle of what this is 50 degree sorry 52 degree and this one is that angle is not given to us let me say this is fine this one is what this is v2 final and this is what v1 final this one is given to us v1 is equal to one meter per second this is m1 m2 we need to find that what is the speed of m2 after cohesion in order to find out it we're going to simple apply what the momentum of conservation in x and y direction along x axis i'll say momentum m1 times 0 .2 plus m2 times 0 here i have i'm going to say that this is v1 this is m1 v1 cause of theta plus m2 v2 cost of 5 now m1 times 2 i mean 0 .2 times 2 which is 0 .4 is equal to m1 which is 0 .2 v1 which is 1 cos of theta this is cost of 52 degree plus m 2 which is about 0 .3 this is v2 cause of 5 from this position i will say that this v2 cost of 5.

03:30 From this position i will say that this v2 cost of 5 is equal to what 0 .905 this is equation first momentum along y -axis along y -axis as we know that momentum along y -j 6 initially will be 0 because they're traveling in x direction what about the this one is like this this is component of the v2 and this is component of v1 in y direction this is v1 sign of 52 degree and this is v1, so v2, cos sine, sign of, as you can see a direction will be opposite to each other.

04:25 So i'll say that m1 v1, sine of 52 degree is just equal to m2 v2 sine of 5...

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