Texts a small ball 0150 kg hanging from a string at a length...

Texts: A small ball (0.150 kg) hanging from a string at a length of 1.19 m is pulled back so the string is completely horizontal. You can assume the radius of the ball is tiny compared to the length of the string. The ball is released from rest and swings down. At the bottom of its swing, the ball collides elastically with a stationary block (1.49 kg). Use the standard coordinate system where up and right are positive. a) What is the velocity of the ball immediately before the collision? b) What is the velocity of the ball immediately after the collision? c) What is the velocity of the block immediately after the collision?

Transcript

00:01 All right, so we have a ball rolling on a frictionless surface, and it collides with another suspended ball like this.

00:09 And it has a velocity va when it does so.

00:12 It's called this ball b.

00:13 So it has a mass mb.

00:16 This is ball a.

00:18 And this ball is on a string of length one meter.

00:23 We're told that va is 10 meters a second.

00:27 The mass of ball a.

00:31 Is 0 .1 kilograms and ball b also has the same mass and so the first part of the question asks what is the speed of the balls just after the collision and we're told the collision is elastic so what we'll have is the momentum of the system initially is just going to be the mass of since both balls have the same mass i'm just going to use the letter m for the mass so the momentum initially is ma and the final momentum is going to be just m times va prime like the momentum after the collision plus m times vb prime and we also know since it's elastic the kinetic energy is conserved so this is one -half mv a squared that's our initial kinetic energy our final kinetic energy is one -half m va prime squared plus one -half mvb prime squared.

01:39 And so if we equate these two, so we'll have like mva equals mva plus mvb prime.

01:50 That's from momentum conservation.

01:51 We can cancel the masses.

01:53 And so we can write va prime for instance as va initial minus vb prime.

02:00 And we can also equate the kinetic energies.

02:02 So we'll end up with the equation looks like this.

02:04 One half mva squared equals one half.

02:07 M v a prime squared plus one half m vb prime squared once again the masses are all going to cancel and the factors of one half we can get rid of those two and so if we substitute this value into this equation what we'll have is v a squared is equal to v a minus vb prime squared plus vb prime so if we expand out the left hand side side.

02:38 We're going to have va squared minus 2 va vb prime plus vb prime squared.

02:47 And then this is plus another term vb prime squared.

02:50 And this is all equal to va squared.

02:53 So we can subtract out those terms...

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