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12) Derive an reaching expression height ` for - the initial h in terms of the velocity of projectilc fired into ballistic pendulum mass of the projectile m1 and bl...

Question

12) Derive an reaching expression height ` for - the initial h in terms of the velocity of projectilc fired into ballistic pendulum mass of the projectile m1 and block mz and and g

12) Derive an reaching expression height ` for - the initial h in terms of the velocity of projectilc fired into ballistic pendulum mass of the projectile m1 and block mz and and g



Answers

A bullet of mass 10 g strikes a ballistic pendulum of mass 2.0 $\mathrm{kg}$ . The center of mass of the pendulum rises a vertical distance
of 12 $\mathrm{cm} .$ Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Okay, so in this problem the trick is this bullet emerges from the block so it doesn't stay embedded in the block. So instead of treating this like any elastic collision, we kind of have to treat it like an elastic one. So the first thing we need to do is to use conservation of energy to figure out how fast the pendulum is going. We can do that because we know the height that it reaches, so the potential energy when it's at its maximum height, it's equal to the kinetic energy just after this bullet leaves the black. And solving for V. Three, we get squared of two G. H. Because the ends are going to cancel Or about 1.53 m/s. Now we can use conservation of the new momentum to figure out the initial speed of the Bullet P one. So the initial momentum and be one is only from the bullet and then it's going to pass through the block. It's momentum will be reduced and it will give some of that momentum to the black. So we can just solve this for everyone. Yet the momentum of the bullet after it leaves the bloc plus the momentum the block afterwards, all divided by the mass of bullet. We get Speed of about 528 m/s.

For this question were given the formula for velocity. Um, with depends on. Well, okay, Sam Capital and G and H. So we're told from the diagram you can see, uh, that the mass of the bullet is 0.0 65. The mass of the block, which is the no, that capital is £6 and the height that the blocks lifts up to is 0.9 feet. And we're also given in The question by G is a constant, which is their duty. So you need to find a velocity of the ball. But 10 years, two feet per second. So all you need to do is what all those numbers into this equation and salt for B. It's a little early to go European 0 65 plus six. And now we're going to divide that by Lord exam, Which is there a 60.0 65. So that part he has us 97 okay, And then now we have to do two times G, which is 32 highs the height of 0.9. Now we're taking back to the power of 1/2 So that is roughly 39. Not only a zero point, not 17 point Ah, 589 And that would multiply that cool 97. We get really ISS roughly. I found it near ST per second, so that would be 736. Treat her second.

In this problem. We have a bullet which is initially traveling to the right in terms of our diagram and strikes, he wouldn't block. Now, being the block swings upward to a maximum angle off six degrees, but initially is at zero degrees. So, using on our information, we want to calculate the initial velocity off the bullet. So the initial system in ST one is the bullet traveling toward the block. And that's the last TV we should calculate for the one. Yeah, The second state is when the bullet strikes, the block and the block bullet system move off together. And the final system at ST Three is when the block reaches its maximum height about the datum and where the data is at the lowest point and that maximum height is 1.25 tens of co sign off theater. Now the first thing we will do is use the conservation of energy and remember, we'll take our data at the lowest point. So the conservation off energy tells us that the kinetic energy off the bullet and block system glass its gravitational potential energy, which is in state to exist after impact. Remember, this is Just as the bullet strikes, the block must equal to the kinetic energy of the system. Last the gravitational potential energy in the final state where the block where the angle theta six degrees. So that's t three less v three. Remember the block there comes to rest. So the initial kinetic energy is a half times the mass off the block and bullet system which is for last 0.0 to that's K g into the velocity of the bullet squid. Knutson state to the kinetic energy since the lowest point, the gravitational potential energy rather is zero since the lowest point is the data and then at Position three when the block moves to its highest position, it has zero kinetic energy since it comes to rest and has gravitational potential energy which is the mass of the system. Four plus 0.0 to K G. Thanks G, which is 9.81 m per square. Second times the height h which is. So this is actually the systems is actually one. This is one minus one. So I see that 11 point 25 minus 1.25 coincide data Yeah, So that's one point 25 multiplied by one minus Coursen feta, which is given to be six degrees so we can see that we have a single equation with a single unknown. So what we calculate here is VB in state two, which is the speed of the bullet just after it impacts the block. And hence we get maybe to be 0.366 five meters the second. But what we actually want is the initial velocity of the bullet. So for the system off, the bullet in the block will now use the conservation off linear momentum reviews. The conservation of energy will now use the conservation of linear momentum and the initial momentum off the block and bullet must equal to the women TEM after impact off the block and bullet system. That's what the new momentum tells us. So initially, the block has no momentum. Since that rest, the bullet has momentum zero point 02 which is its mess times its initial velocity V B one, and this was equal to the momentum of the system after the initial impact, which is for plus 0.2 as the blocking. Will it move off together and they move off together with velocity, We calculated to be zero point the 66 five So we can rearrange this and find the initial velocity of the bullet as required, maybe one to be 733 meters the second, and that's to the right.

So we have uh we have to prove that the initial velocity of project type is given by the equation we notice equal to M plus M upon um multiplied by toys. G. Yes. Uh So uh first let's apply the energy conservation law for the collision of bullet ballot. Nd bob. So um that is half an plus in the square plus and plus M G H one is equal to half M. Plus you square. Yes. Um blast em G. And this situation one here. Um and the equation for momentum conservation for the collision of bullet and the bobby's M. We note plus M. P. I. Is equal to M plus M. V. This is equation two. So now first let's put the values in equation one. So we have half M plus M. P square air plus M plus M G H. One is zero m is equal to half M plus m zero m per second plus M plus M gee. It's that is equal to have M. Plus M. V squared is equal to M plus M G H. From here we'll get the is equal to route over twice. And she edge. Now let's put the value of V I in equation to there is um we not plus M V Y is V R E zero m per second. Well um plus I am the so from here we're not is equal to M plus them. Now let's put the value of V from here. So we have we can say we notice equal to M plus M upon em multiplied by the east to route over twice G. Which therefore um it is proved that the initial velocity of projectile is M plus M upon them, multiplied Beirut or twice she hands It is shoot.


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