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A 12.0 -g rifle bullet is fired with a speed of 380 $\mathrm{m} / \mathrm{s}$ into a ballistic pendulum with maxs 6.00 $\mathrm{kg}$ , suspended from a cord 70.0 $\mathrm{cm}$ long (see Example 8.8 in Section 8.3 . Compute (a) the vertical height through which the pendulum rises, (b) the initial kinetic energy of the bullet, and (c) the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendalum.

a) $h=2.9 \mathrm{cm}$b) $K_{b}=866.4 \mathrm{J}$c) $K_{t o t}=1.73 \mathrm{J}$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

Cornell University

University of Michigan - Ann Arbor

University of Washington

Simon Fraser University

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In this question, a bullet off 12 g is fired at a velocity of 380 m per second in the direction of a ballistic panned along with the mass off 6 kg and a length off 70 centimeters. Then that bullet will penetrate in the ballistic pendulum and after that the pendulum will swing, reaching some height age. Then in the first item off this question, we have to better mind. What is that? Maximum hate age. For that we can use energy conservation. But be careful because energy conservation holds in between two and three Onley. Because during this collision some energy is lost in this process. The reform energy is not conserved in between one and two. It is not destroy it. Also, it actually becomes heat the reform in between one and two, some energy is transformed into heat. So we can't use the principal off energy conservation between processes one and to the reform. What we can do is the following in between two and three energy is conserved. It means that the total energy e in situation number two must be equal to the total energy e in situation number three in the situation Number two we Onley have the kinetic energy off the ballistic pendulum and the ballot. This is provided that we choose a reference frame such that our horizontal axis crosses right here We're appointing. So in fact, this position means that the Y coordinate or the height in some sense is equals to zero. Therefore, there is no gravitational potential energy Given that we choose this reference frame which is the frame that I'm choosing the reform in situation number two The energy is given on Lee by that kinetic energy. So we have the mass off that bullet plus the maths off. The pendulum on this is multiplied by the velocity off the set we squared and then divided by two. That velocity v squared is the velocity with which both the pendulum on the bullet are moving after the collision. Then we read situation number three In situation number three, the pendulum will swing on Then it will stop at some height on swing back at that maximum height, the pendulum has zero velocity to reform. There is no kinetic energy. When the pendulum reaches the maximum height, then all the energy in situation number three is gravitational potential energy and the equation for a gravitational potential energy is the following. It is given by the mass actually the mass off the bullet plus the mass off the pendulum because both are at that height, age, times G, which is the acceleration of gravity times the height age. Now we solve this equation for the height age. But before let us simplify out a common factor. We can divide both sides off this equation by that factor. So we end up with the following relation. We squared over two is equal to g times age, then h the maximum height is equals to the squared divided by two times G. Okay, but we don't know what is that Philosophy V. How can we proceed? Well, we can actually discover that velocity v by using the principle off mo mentum conservation in between situations one and two. Because even though energy isn't conserved from one to chew, the momentum must be conserved. So we can say that the net momentum in situation number one soap you not one must be equal to the net momentum in situation number two, which is pure net to in situation number one. All the momentum is coming from that budget. Moving the reform, the net momentum Number one is equal to the mass off the bullet times the initial velocity off the bullet. After that collision, both will be moving with the same velocity which I'm calling the V. The reform The net momentum after the collision. So the net momentum situation number two is given by the total mass M B plus M p, which is the mass off the pendulum times the final velocity V. Then we can solve this equation for big V. So big V is equal to the mass off the bullet times the initial velocity off the bullet divided by the mass off the bullet plus the mass off the pendulum. Now that we know what is big V, we can substitute this result into this equation and by doing that, we get the following the maximum height Age is given by big V squared that is M b times VB divided by m b plus M. P. This is squared times one divided by two times G. Now we plug in the values. So by plugging in the values we get the following the mass off the bullet is 12 g 12 g is the same as 12 times 10 to minus tree kilograms. Remember, from grams to kilograms, all we have to do is divide by 1000 or multiplied by 10 to ministry which has the same effect. Then we multiply this by the initial velocity of the bullet which is 380 m per second. Now we divide this by the mass off the bullet 12 times 10 to minus three, plus the mass off the pendulum which is 6 kg. Then we multiply it by one, divided by two times the acceleration off gravity which is approximately 9.80. On these results in the maximum height that is given by approximately 0.294 m, then converting these quantity from meters two centimeters we get that the height is given by 2.94 centimeters approximately. Remember from meters, two centimeters we multiply by 100 on. This was the first item off this question For the second item, I you need some space, so I will reorganize my board. Okay. In item be All we have to do is evaluate the initial kinetic energy off. The bullet on this is as simple as doing the mass off the bullet times the initial velocity of the bullet squared divided by true. This results in 12 times 10 to minus three, which is the mass off the bullet in kilograms times the initial velocity off the bullet 380 squared, divided by two. These results in the initial kinetic energy that is approximately 866. Jules, this is a very respectable energy. Now, in the next item we have to evaluate what is the kinetic energy off the bullet on the pendulum here in situation number two, right after the collision again, This is a simple was doing. The following K two is equals to the mass off the bullet embi plus the mass off the pendulum MP times the velocity off the bullet and the pendulum squared divided by two. We know that the velocity off the bullet and the pendulum is given by his expression. So K two is actually given by M B plus MP divided by true times M b times VB divided by m B plus MP on. This is squared as you can see, we can simplify one factor off EMBI plus MP with one factor off Embi plus MP that will come from this term which is squared. By doing that, we get the following Que true is equals to m B squared times V B squared divided by two times and be plus m p. Now we just have to plug in the values and finish the question So M B is 12 times 10 to minus three. This is squared. Then we multiply it by VB which is 380 squared and divide by two times m b which is 12 times 10 to minus three plus MP the mass off the pendulum which is 6 kg. These results in my kinetic energy Ke chew that is approximately 1.73 jewels. So as you can see ah, lot off energy was lost during this process and that energy loss happens by heat and also by deforming the pendulum and they use is the answer to this question

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