(I) In a ballistic pendulum experiment, projectile 1 results i...

Question

(I) 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, h₂ = 5.2 cm . The second projectile was how many times faster than the first?

Accepted Answer

Welcome back. Everyone. In this problem, we want to calculate the speed of the second ball relative to the first during our collision pendulum experiment where the first projectile ball elevates the pendulum to a maximum height of three centimeters. And another projectile ball of the same mass triples the height to nine centimeters. For our answer choices. A says that it's the square root of two multiplied by V one B says it's the square root of three multiplied by V one C, the square root of six multiplied by V one and D, the square root of seven multiplied by V one. Now, let's just make a note of everything we know and what we want to know here. So we know, OK, we have two speeds. V one and V two. We don't know what they are. OK. But we know that our first projectile ball V one, sorry. Our first projectile ball elevates the pendulum to a maximum height of three centimeters, which tells us then that H one equals three centimeters. And we also know that our second ball elevates the height to nine centimeters. So in other words, H two equals nine centimeters and in this case, we want to calculate the speed of the second bar. In other words, when we look at our answers, we want to express V two in terms of V one using the information that we are given. Now first, let's write the formula to find the initial projectile speed. So for our projectile speed in our comparison, we know, OK, we know that V is going to be equal to M plus MK divided by mm multiplied by the square root of two G where V is the initial projectile speed, M common M and capital M are the masses of the 1st and 2nd projectiles respectively. G is the acceleration due to gravity and H is the height here. The masses of the two projectiles are the same. So if we want to express V two in terms of V one, let's compare their speeds. OK. So comparing their speeds, comparing their speeds, then we can apply the formula to V two divided by V one. We're comparing both of their speeds here. So taking our formula, OK. Then that means it's going to be equal to for V two M plus M divided by a multiplied by the square root of two GH two. OK. Because their masses are the same divided by V one, which the expression would be M plus M divided by M multiplied by the square root of two GH one. Now, what do we notice here? Well, here we know the expression M plus M divided by M all cancels. And that leaves us with the square root of two GH two divided by the square root of two GH one, which we can rewrite the entire quotient under the square root. So that's two GH two divided by two GH one and no two G is cancel the levers with V two divided by V one being equal to H two divided by H one. Now that we have this formula we can solve for V two if we multiply both sides by V one, that means V two equals V one multiplied by H two divided by H one. No, so far. Oh, sorry. The square root rather, I forgot my square roots and my apologies because the square root itself didn't cancel, right. It was just the, the terms under the square root. So now we have an expression for V two in terms of V one and we know what H two and H one are. We already know that H two is nine centimeters and that H one is three centimeters and no. OK. So all of that is multiplying V one. So that means V two is going to be equal to V one multiplied by nine divided by three, which is the square root of three or is going to be equal to the square root of three multiplied by V one. OK. So the speed of the second ball relative to the first is the square root of three multiplied by V one. If we look back at our answer choices, that means B is the correct answer. Thanks for watching everyone. I hope this video helped.

Struggling with Physics?

(I) 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, h₂ = 5.2 cm . The second projectile was how many times faster than the first?

Watch next

Ballistic Pendulum practice set