A 0.280-kg croquet ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Before the collision of these crochet balls, we have the first one coming in at some unknown velocity and the other crochet ball is initially at rest so v 2 is zero and the crochet ball one has a mass of 0.280 kilograms. We don't know what the mass of ball two is and that's what we have to figure out in part (a) and we also know that the velocity of this ball two after the collision so v 2 prime is gonna be the velocity of the first ball, v 1, divided by 2. So velocity of the first ball before the collision divided by 2 and the velocity of the first ball after collision we don't know and also don't care about because we don't have to calculate it. So we are gonna write down the things that we know in formula form here. So we have conservation of momentum and we have this piece of information that we are given that the crochet ball two after collision is gonna have a velocity half that of the first ball before the collision and because the collision is elastic, we know that v 1 minus v 2 equals v 2 prime minus v 1 prime. Now v 2 is zero because this crochet ball is initially at rest and so these two terms are gone and we can rewrite equation 1, version b by getting rid of that term with v 2 factor in it and then also substituting for v 2 prime. So instead of writing v 2 prime, I'm gonna write v 1 over 2 in its place and otherwise, this is copying this formula here while omitting that term. And then equation 3, we are gonna get rid of this v 2 term as well minus v 2—that's just zero there— and so I copied down just v 1 and then equals v 1 over 2 instead of v 2 prime and minus v 1 prime and we are gonna solve this equation 3 all the way until we get v 1 prime equals. So this is v 1 prime minus v 1 or we can take this term to the left hand side I guess that's one way to look at it and that gives us positive v 1 prime and then we'll take this to the right hand side which makes this v 1 negative. So we have v 1 prime is v 1 over 2 minus v 1 and multiply this by 2 over 2 to give these two terms common denominator and you have 1 minus 2 is negative 1 and so this negative one-half v 1 is gonna be v 1 prime. So we'll substitute for that in later calculations here or the other formula. So it's kind of interesting though we didn't have to calculate what the velocity of ball one would be after the collision but we ended up doing it anyway. It's gonna be half of its original velocity but in the opposite direction so it's gonna be rebounding to the left so I guess I should put the arrow to the left; it wasn't possible to really tell until we did the algebra that it should go to the left. So substitute this result into equation 1, version b and we are gonna replace the v 1 prime with negative v 1 over 2. And so we have m 1v 1 equals m 1 times negative v 1 over 2 plus m 2 times v 1 over 2—let's just copy it from here— and we can divide each term by v 1; that's convenient, it's nice that that happened that this is a common factor for every term so divide both sides by it and it cancels out and we have m 1 equals negative m 1 over 2 plus m 2 over 2. Multiply everything by 2 to get rid of that denominator and that gives 2m 1 equals negative m 1 plus m 2; bring this to the left hand side by adding it to both sides gives us 3m 1 and switch the sides around so we have the unknown m 2 on the left and we have m 2 is 3 times m 1 so that's 3 times 0.280 kilograms and that gives us 0.840 kilograms. And the next question is what fraction of kinetic energy is transferred to ball 2? So the amount of kinetic energy transferred to ball 2 is gonna be its final kinetic energy minus its initial kinetic energy but it has no initial kinetic energy and so this change will be just the final kinetic energy of ball 2 since this term is zero and that's one-half m 2 v 2 prime squared. But v 2 prime is v 1 over 2 as we know from this piece of information there and so when we plug that in, we end up with kinetic energy of ball 2 is m 2v 1 squared over 8; this 2 in the denominator got squared to make 4 multiplied by this 2 makes 8. And then the total initial kinetic energy of this system is only that of ball one because ball two is at rest and so the initial kinetic energy is one-half m 1v 1 squared and so when you divide these, you have this term m 2v 1 squared over 8 divided by this other thing but instead of dividing, we are going to multiply by its reciprocal because dividing fractions by fractions is confusing so let's multiply by reciprocals instead. And so we are gonna multiply by 2 over m 1v 1 squared and well these things cancel—that's nice— and we have 2 over 8 which reduces to one-quarter. So one-quarter of the kinetic energy is transferred to ball two after the collision.