🎉 Announcing Numerade's $26M Series A, led by IDG Capital!Read how Numerade will revolutionize STEM Learning

Like

Report

Numerade Educator

Like

Report

Problem 80 Easy Difficulty

A $20.00-\mathrm{kg}$ lead sphere is hanging from a hook by a thin wire 3.50 $\mathrm{m}$ long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a $5.00-\mathrm{kg}$ steel dart that embeds itself in the lead sphere. What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision?

Answer

$v_{1}=65.5 \mathrm{m} / \mathrm{s}$

Discussion

You must be signed in to discuss.

Video Transcript

{'transcript': "so exactly size We have a sphere off mass M A that is hanging in a wire off length are as initially, this fear is stationary and this year is about to get hit by a particle off mass M B on the velocity V b. So, after getting hit by the particle and be the particle, uh, it's text together with the sphere forming a single body off total mess M A M B that will now gain a velocity V chew. And due to this velocity, the the body M A and B in blue will be allowed to rotate in the form of a circle of radius R. And we want to know what has to be the minimum velocity V B The initial velocity off the particle be such that the body and may and be will complete a full circle like this. Okay, so that that should line to be a trajectory off the body. M A and B. So I want to find this minimum velocity. Okay, So, um, to start solving this exercise, we can, uh, first we can look at the conservations that can be use it. So let's look at scenario wants to senator to you. So scenario 12 scenario to you What kind of conservation we have. So, um, besides, off the gravitational force, there is no external force. And the gravitational force is only pointing to the Y direction. So in the X direction, there is no force acting on it. And since developed, we know that the velocity of particle we is pointing to the X direction we can have right before the collision and right after the collision. Conservation off the linear momentum in the X direction. So right here, Lena P X. But notice that we cannot use conservation off energy from scenario wanted to because the particle a get sticks together with the sphere. Sorry. The particle be it sticks together with the sphere eight. And this requires a and for the bodies A and B to be together. Some off the mechanical energy on the scenario. One has to be lost in the former heat, uh, to get to scenario too. So from scenario 1 to 2, we have conservation off Lena Momentum. So let's, uh let's see what the conservation of nongovernmental gets us. Sorry, it gives us so Lena romance. Um so good here. So first, let's apply conservation offline. In moments of form scenario one has to be equal to the linear momentum off scenario, too. In scenario one, we only have the particle off mass M B moving with velocity V B, which is the quantity we want to find in the end off the exercise. And in scenario two, we have the full particle M A plus M b traveling with initial velocity Connecticut. We called the h of Okay, I'll call it. Sorry. Just meet you. So this with feature and we have a relation between, uh so from the conservation of the momentum, we have this relation between the initial velocity off particle be with the initial velocity off the off the body a plus B, which is equal to I am a plus m b divided by m b times v chew. So this is what conservation of leading moments gives us. Now let's go back to our sketch. So we have that from scenario to two scenario three. No mechanical energy is lost. And so we have conservation off mechanical energy. So put here chemical energy. Okay, but we don't, But we cannot imply incarceration off linear momentum Because, uh, from the moment that the body is in the bottom to the moment that the body reaches the top, we have the action off the gravitational force and the action off the attention off the body. So we have. So we will lose some, uh, lena momento due to the external forces during the whole time it goes from the bottom to the top off the trajectory. Okay, so that reported here, too. We have conservation, uh, mechanical energy. So energy between scenarios to two scenario between number two and three. So we have that the mechanical energy off scenario, too. So the kinetic energy off center two plus the, uh, but gravitational energy on scenario to has to be equal to the kind of energy on scenario three plus the gravitational potential off scenario three. Okay. So we can always choose where our zero off the gravitational potential ISS. And so to do so because of this, I will choose it to be down here in the initial position off body A plus B just thio make the calculations easier. Okay, such that, um, here it will be zero. And we only have the kinetic energy. So Scenario two is the mass off particle way plus the mass of particle be times V two squared over to and this has to be equal. Sorry, this has to be equal to massive particle. A plus mass off particle beam. The three squared over two. So the velocity off body A plus B in the top of the trajectory, plus the gravitational potential. So let me write it down here. Plus mm A plus m b times g times the height off the trajectory. So since we chose our zero to be here the height, the relative height between the zero to the current point, it will be two times the rage is off the trajectory. So since it's going to be here, the ball, the height that will use to calculate the gravitational potential on the top will be, uh, to our so times to our okay, we can divide both sides of the equation by m a plasm be okay. So we have that so that we'll have the relation between the velocity at the bottom of the trajectory to the velocity at the top of the trajectory. So teach you squared over two is equal to V three squared over two plus g times to our okay. Now, let's keep this. So I'll call this relationship the previous relation. All college relation one. So remember that we want to find VB. So to find Phoebe, you must find with you. But to find with you we have to find the three. So how do we find victory so we can find it from, uh, Newton's second law. So here, Newton's second law. So we have a body that is moving in a circular orbit. So we have that, uh, m the mass off the body times the radio acceleration. Call it a R is going to be. So In our case, the mass is m A plus M b. This is going to be equal to sorry. This times the accelerate the radio acceleration, which is equal to the, uh, tangential velocity squared over the radius off the off the orbit. Okay. And this is going to be equal to the net force on the system. So, um, if we pick the scenario three on the way to the particle, almost to the body is at the top of the trajectory. We have that Both the gravitational force is pointing downwards, and the tension on the wire is going is pointing outwards. So we have that RV is going to be be three on its minus T plus, um, the gravitational force, which is I am a plus m b times G. Okay, now is when we use the condition off the minimum velocity required to the body. And may plasm be complete a loop in the X Y plane. Sorry. So we have that The minimum velocity, uh, is going to be the velocity on which to n the body A plus B reaches the top of the trajectory. The tension zero. So this is the condition. Okay. So because if the tension is greater than zero than the velocity will have to be greater to compensate the total downward, uh, force forcing the body to go to the bottom. Okay, so we'll have that. If the force in the bottom is is great, it's too big. The velocity will have to be even greater to compensate this downward downward movement for from the, uh, net force. So if the tension is zero that the force that will be pointing that words will be smaller. So the velocity required to compensate the gravitational force will be smaller. Okay, Will be the minimum velocity. Hope it's not confusing. So, uh, we just replaced this in our equation. And we have that Vetri So have m a plus and be the tree squared over r This is going to be equal to. So I'll just put the mind off aside because this minus sign is just specifying us that the acceleration is pointing downwards to the center off the circle. Okay, so this is going to be equal to, um a plus on B tonic G. Okay. So we can cancel out these terms and we have that The velocity off the Sorry. So the velocity off the particle off the body a plus B at the highest point, the highest point of the trajectory is going to be equal to the square. Root off the Raiders of the trajectory times the gravitational force, the gravitation acceleration. Sorry. And so we stood in the values that we have. We find this value to be equal to 13.1 meters per second. Okay? Or call this tree. So Okay, so we know the value off the velocity on points on the highest point of the trajectory. So we can use it to calculate the initial velocity off body A plus B to them. Calculate the velocity off the particle, be right before it hits the sphere off mass A off the fear of mask and A Okay, so let's be stood three and two. And by doing this restitution, we find that Oops. Sorry V three for R equals to 3.5 and J equals to 9.81 p. 13.1 is going to be equal to 5.84. Okay, sorry about this. Okay, so now to be student is, uh we put a precise value here, So if you simply stood this in expression to we find that V two is going to be equal to 13.1 m spare second. Now, we should be stood this into expression one from the conservation off linear momentum. And we have that VB okay is going to be equal to the mass off the sphere. Eso 20 kg plus the mass off the particle be which is equal to 5 kg over the mass of particle being times 13.1. Okay. And this is equal to 65.5 m per second. So this is the minimum velocity the park will be has to have in order to the body a plus B complete full up in the X Y plane."}