In a ballistic pendulum experiment, projectile 1 results in a maximum height $h$ of the pendulum equal to 2.6 cm. A second projectile (of the same mass) causes the pendulum to swing twice as high, $h_2 = 5.2 \textrm{ cm}$. The second projectile was how many times faster than the first?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. We need to come up with some relationship between the height of the pendulum goes up to and the initial velocity of the bullet so that we can figure out by what factor the velocities are different between these two bullets based on knowing how the heights are different that the ballistic pendulum goes up to. So just after the bullet embeds in the pendulum, there's conservation of momentum; there's no conservation of kinetic energy because the collision is inelastic which means kinetic energy is not conserved but momentum is still conserved so we can use this here. We can say that the momentum of the system which is just that of the bullet since the pendulum is at rest initially is m times v and then after the collision, the momentum is gonna be, well, the total mass of the bullet embedded within the pendulum times their common v prime velocity that they have after the collision. So this is just after the collision, before there's any rise in the pendulum because as soon as the pendulum starts to move upwards then it's working against gravity and so that's an external force on the system in which case momentum is not conserved when there's an external force involved but this is just for that instant when the bullet is embedding itself in the pendulum. So that's all we can say about that and then we can say a while after the collision, we have that the potential energy when the ballistic pendulum is at the top of its swing equals the kinetic energy that it has just after the bullet is embedded. So this initial moment is just after the collision is complete and the embedding of the bullet is complete and at this point, it has a velocity of v prime squared and well, times m plus M times a half to get the kinetic energy and that's gonna be the potential energy at the top of its swing the total mass m plus M times g times the height, h. And we can take equation 1, solve it for v prime by dividing both sides by m plus M and we have the v prime is little mv over the sum of the masses and then we'll substitute that into this conservation of energy formula here and figure out how the initial velocity when the bullet is approaching the pendulum is connected to the height the pendulum rises to. So we substitute in for v prime by writing its mv over m plus M and that's gonna be squared and otherwise, this is just a conservation of energy formula copied. In case you are a little bit confused why I'm talking about conservation of energy in a problem which has an inelastic collision during which kinetic energy is not conserved, it's because we are considering two different moments separately so we have this conservation of energy does apply after the bullet is completely embedded and this pendulum is starting to freely swing. The total kinetic energy that it has, whatever is left over, I mean there's some kinetic energy that the bullet has initially, and some of that gonna's be lost because the collision is inelastic but once that's all taken care of and we have some final well, I'm using the word 'final' a bit confusingly but we have some kinetic energy leftover in the bullet and pendulum combined after the collision is done and that kinetic energy is conserved as the pendulum is swinging upwards. Hopefully that makes sense. So we have a common factor on both sides—m plus M— and that cancels and we multiply both sides by 2 and multiply both sides by m plus M squared and then divide both sides by m and we end up with this line here, after switching the sides around, we have that v squared— initial speed of the bullet as it's approaching the pendulum— equals the sum of the masses squared divided by bullet's mass squared times 2 times g times the maximum height the pendulum reaches. And then take the square root of both sides to solve for the velocity of the bullet before hitting the pendulum and that equals m plus M divided by m times square root of 2 times acceleration due to gravity times the height of the pendulum. So this is true for both bullets and so we substitute h 1 in here for v 1 and h 2 here for v 2 because otherwise we are told that the bullet and the pendulum mass is the same. And so we are trying to find by what factor does v 2 differ from v 1? And so let's figure out what x is; x is gonna be v 2 divided by v 1 and v 2 is m plus M divided by m times square root of 2 times g times h 2 and then dividing by v 1 is better done as multiplying by its reciprocal so we are gonna multiply by m over m plus M times square root 2gh 1 and there's a lot of common factors here; this cancels, this cancels and the square root 2g cancels from both top and bottom as well and so this factor is gonna be square root h 2 over h 1. So that's gonna be square root of 5.2 divided by 2.6 and there's no need to convert these into meters because the units are just going to cancel anyway, the only thing that's important is that their units are the same and this means we get 1.4. So velocity 2 is gonna to be 1.4 times faster than velocity 1 given these heights of 5.2 centimeters versus 2.6 centimeters.