Please answer and solution if needed all that following...

- An object weighing 50 kg was dropped from a height of 10 m. What was its momentum the instant it struck the ground? What was its K.E.? Neglect air resistance. - A man claims that he can hold onto a 12.0-kg child in a head-on collision as long as he has his seat belt on. Consider this man in a collision in which he is in one of two identical cars each traveling toward the other at 60.0 mi/h relative to the ground. The car in which he rides is brought to rest in 0.10 s. Find the magnitude of the average force needed to hold onto the child. - Two practically elastic spheres collide. If their velocities before impact have same magnitude and opposite directions and one has a mass 3 times the other, what are their velocities after impact? - Charles throws a basketball vertically upward at a speed of 5 m/s. The ball reached the maximum height then drops freely. If Charles fail to catch the ball on its way back down, find the height of bounce of the ball (assume the coefficient of restitution between the ball and ground be 0.6).

Transcript

00:01 In this problem we have two spheres with an initial velocity colliding off each other in all three physical dimensions.

00:14 We've been test defining the final velocity of one of the spheres after the collision.

00:22 So let's start with what we know.

00:23 We know the masses are going to be constant, m1 and m2.

00:27 M1 is 0 .5 kilograms, and m2.

00:33 Is 1 .5 kilograms.

00:38 Now we don't know what type of collision this is, so we're going to just use the standard conservation of linear momentum equation.

00:48 To do that, it would just be m1v1 initial plus m2v2 initial equals to m1v1 final plus m2 v2 final.

01:08 For part a we're going to find the velocity if we know the final speed of one of the balls.

01:15 Let's put everything we know into this equation.

01:19 So mass of 1 is 0 .5.

01:21 The initial velocity of 1 is 2i minus 3j plus 1k.

01:35 The mass ball 2 is 1 .5 and the initial momentum of initial speed of that is negative 1i plus 2j minus 3k.

01:50 So that's our initial momentum of each ball and this is going to equal to the final momentum of each ball.

01:57 Now we only know the velocity of one of these.

02:03 So let's figure out what it's going to be.

02:06 We only know the velocity of the mass one ball.

02:11 So 0 .5 multiplied by vf which is negative 1i plus 3j minus a k and that's plus 1 .5 v2f.

02:31 So from here what we can do is expand out all of our parentheses and figure out how many i's jays and k's we have on both sides so let's do that now.

02:42 From the first one we have one i minus 1 .5j plus 0 .5k minus 1 .5 i because of that negative 1 .5 j.

03:01 Plus 3j minus 4 .5k which is equal to negative 0 .5 i plus 1 .5j minus 4k plus the 1 .5v2.

03:33 Then by combining like terms and rearranging we end up with, well we'll combine like terms first.

03:45 Negative 0 .5i plus 1 .5j minus 4k is equal to 1 .5vv2f and then the same thing so minus 0 .5i plus 1 .5j minus 4k.

04:15 So we'll bring these three onto the other side of our equal.

04:22 Sign and actually what you should see is that they are both the same so v2f is actually equal to zero meters per second and for it to be zero meters per second it has to be an elastic because they're most moving at the same initial or same final speed vf okay part b we have a similar situation now but we have a different final speed so i'm gonna use what we ended up with in the first part of the equation in part a and has negative 0 .5i because our initial speeds are the same plus 1 .5j minus 4k is equal to 0 .5 multiplied by its new final speed which is now assumed to be different that is negative 0 .25i plus 0 .75j minus 2k plus 1 .5 v2f.

05:52 So in the very similar situation to what we were in just a moment ago.

05:56 Again, i'm going to expand these parentheses on the right hand side, and then we'll rearrange.

06:02 So, expanding that out, gives us negative 0 .125 i plus 0 .375j plus 0 .375j.

06:18 Minus 1k plus 1 .5 v2f.

06:25 And now we'll rearrange, we'll bring everything on this side, apart from the v2f, over by adding or subtracting.

06:36 So what we should have is negative 0 .375i plus 1 .125j plus 1 .125j minus 1 .15j, minus 375i, plus 1 .125j, 3k is equal to the 1 .5 v2f.

06:59 Then if we divide both sides by 1 .5, what we should get for v2f, hopefully is negative 0 .25i plus 0 .75j minus 2k meters per second.

07:33 So that's a final speed of ball 2 with a different set of final speeds for ball 1.

07:40 Okay, now, part c, again very similar, except now we have an unknown variable in the k direction, as well as an unknown final velocity for the second ball with the mass of 1 .5.

07:56 I'm going to start off again with the similar equations we had in the last question, negative 0 .5 .5.

08:04 Plus 1 .5j minus 4k.

08:12 That's our initial total momentum, again, the same thing.

08:19 And that's equal to 0 .5, multiplied by negative 1i plus 3j, plus a, k, plus our, again, unknown, 1 .5v2f.

08:43 So i'll expand the parentheses here.

08:46 We end up with negative 0 .5i hat and positive 1 .5j hat and 0 .5a k hat plus 1 .5vv.

09:07 Now what we should see here is that 0 .5i is negative on both sides and positive 1 .5j is there on.

09:17 Both sides as well.

09:19 So what they do is go away and we end up with negative 4 k hat is equal to 0 .5 a k hat plus 1 .5 vf.

09:34 So i want to find out what a is equal to here so i can just rearrange that real quick and knowing that the final velocity has to be going in the k direction, that's going to help me...

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