I am stuck by these 2 problems. Can someone teach me how to do these questions? I would really appreciate! Please Register or Log in to view the hidden image! 1) A cannon of mass 1000kg is mounted on wheels free to move on a horizontal surface. It fires a 10kg cannon ball which leaves the cannon travelling 300m/s at an angle 15 degrees above horizontal. Calculate the 'recoil' velocity of the cannon. [Can I solve this problem using 1-D methods? Will the cannon have a velocity in the downward component?] 2) A 4kg block is suspended from a spring with a spring constant of 500N/m. A 50g bullet is fired into the block from immediately below with a speed of 150m/s just at impact. The bullet becomes embedded in the block. 2a) Determine the amplitude of the resulting SHM. [To find the amplitude, should I treat the equilbrium position as having kinetic energy or not? Because the question implys the block is at rest before the bullet is shot, is initial velocity of the block after collision still zero at the equilibrium position?] 2b) What fraction of the original kinetic energy of the bullet is transferred into mechanical energy of the harmonic oscillator? [I got Ek of bullet=562.5J and total mechanical energy of the block+embedded bullet = 8.4810706J and 8.4810706/562.5 = 1.51%. However, the answer says it's 1.23%. Am I doing something wrong?]
Hi, #1: Momentum can be split into components, just like velocity. In the vertical direction, the Earth's momentum will be affected. The Earth will also recoil. #2a: If the velocity of the block after the collision were zero, the thing wouldn't move and there'd be no amplitude. #2b: The books answer is correct. I don't see how you got the total mechanical energy as 8.48J. How'd you work it out?
HI, 1) How can I take care of the vertical component of the momentum? Will the cannon's 'recoil' velocity be completely horizontal? If not, how can I find out the magnitude and direction of the velocity after the interaction? Will the cannon crash or sink into the ground? 2a) So you mean that the initial velocity at the unstretched position is not zero... Will the block be displaced DURING the collision? If so, how do I know at which position the combined mass will be AFTER the collision? DURING the collision, the block gains acceleration, so I think the block will move during this time and this makes thing very complicated..... 2b) At the equilibrium position, the box's total mechaincal energy is elastic potential energy + kinetic energy, right? I just substituted in the values... I will go and check my work now... Please Register or Log in to view the hidden image!
#1) You're probably not expected to take into account the vertical component of the momentum. If you want to, you can work out at what speed the 6 sextillion tonne Earth recoils in the vertical direction (M<sub>Earth</sub> = 5.97x10<sup>24</sup> kg). #2a) You could just make life easy on yourself, and assume the collision was instantaneous. #2b) At all positions, the mechanical energy is KE + ΔPE. You should use the equilibrium as your reference point (so the PE there is zero).
1) Can we just consider the cannon and the cannon ball for this problem? Why do we have to care about the Earth? 2b) Must we pick the reference point for gravitational potential energy at the equilibrium position? Can we pick any other position?
1) In the horizontal direction, you don't need to take the Earth into account. In the vertical direction, you do because of the normal force between the cannon and the Earth. If you actually bother to do the calculations for the vertical component, you'll get such a tiny speed its not worth mentioning. 2) Why would you mant to do that? You might as well set the position of minimum PE as zero. I'm not sure how mechanical energy is defined, but it makes no sense for it to depend on where you felt like setting your reference point. And setting the equilibrium as zero PE gets you the answer in the book (and simplifies the calculation you have to do).
1) So for the ball-cannon 2 object system, is momentum only conserved only in the horizontal direction? (because no external horizontal net force) 2) Total mechanical energy of the block+embedded bullet = Elastic energy + kinetic energy + gravitational potential energy At the beginning (just after the impact), the combined mass have kinetic energy. It also have elastic energy because for a mass hung on a vertical spring, it will drop and reach a new equilibrium position that is not unstretched. We define x=0 be the unstretched position, i.e. elastic enrgy=0 beucause no elastic energy is stored. I indeed got the answer of 8.4810706J......
#1: Yes. #2: You define the equilibrium position (also the position of maximum KE) as having no potential energy of any kind - ie take it as the reference point for both the elastic and gravitational PE (Remember how the unstretched position doesn't have much significance in a vertical spring?).
1) According to the law of conservation of momentum, total momentum is conserved if the net force acting on a system of interacting objects is zero. In this case, in the vertical direction, even if there is a large normal force acting on the cannon, the vertical net force is still zero, right? Shouldn't momentum of this 2 object system also be conserved in the vertical direction, then? 2a) Can part a) also be solved by setting x=0 as the equilibrium position instead of the unstretched position? Will I get the same answer? The way I have previously done this question is by setting x=0 as the unstrecthed position. I first find the new equilibrium position below the unstretched position after the object is hung on the vertical spring...and then use the law of conservation and energy respectively. I got the correct answer this way.
I have a few more conceptial problems that I hope I can understand! :m: 3) A small truck and a large truck have the same kinetic energies. Which truck has the greater momentum? [I have tried writing out the expressions, but it seems that there are too many unknowns to solve this problem...] 4) A moving object collides with a stationary object, is it possible for both objects to be at rest after the collision? 5) A wet snowball of mass m, travelling at a speed v, strikes a tree. It sticks to the tree and stops. Does this example violate the law of conservation of momentum? [I don't think the law is violated. But what is the external net force in this case that acts on this snowball-tree system, making the net force of the system non-zero?]
Only if you consider the Earth to be a part of your system. If you do, you have to take the Earth into account in your momentum calculations. If you define your system as just the cannon and cannon ball, then the force the Earth applies on the cannon is an external force, as the reaction force acts on something that isn't part of the system. Where you set x=0 shouldn't affect your answer to 2a).
Hint: Kinetic energy = momentum * velocity (If you're not too fussy about multiplying vectors). What is the total momentum before the collision? What is the total momentum after the collision, if both objects are stationary? Does the tree float in mid-air?
Er, no it doesn't. Kinetic energy is K=(1/2)mv<sup>2</sup>, whereas momentum is p=mv. The correct relationship between kinetic energy and momentum is K=p<sup>2</sup>/2m
Oops :m: It was, uh, a test to see if anyone would notice Please Register or Log in to view the hidden image! Or alternatively, K = ½pv (=½(mv)*v). I just left out the ½. But I agree that K = ½p²/m is more useful for kingwinner's problem. I'm not sure why I suggested the other one.
4) So the answer is "impossible if frictionless". What if there's friction? Will it be possible for both objects to be at rest after the collision? 5) I don't get why the tree-snowball system have a non-zero net force. And what is the force that causes this net force of the system to be non-zero?
Depends on the friction forces involved. Technically, I think it'll always be zero anyway (the two objects will drag whatever third object is causing the friction). System consists of: 1 snowball and 1 tree. #1 Does the tree move or not when the snowball hits it? #2 If it doesn't move, why not? What could supply a force on the tree such that the net force on it is zero?
Hi, 4) If the floor has friction, and when a moving object collides with a stationary object, is it possible for both objects to be at rest after the collision? 5) The tree doesn't move when the snowball hits it. Because the ground applies a force on the tree, I think. And prehaps this is the external net force that make the snowball-tree system to have a non-zero net force. Thus, momentum is not conserved in the snowball-tree system... :m:
Relative to floor, yes, but who says the floor is still moving at the same speed it was before? Imagine the collision taking place on a table in space (this table is heavy enough to attract the objects gravitationally). The reaction to the friction force will cause the table to accelerate. The earth is just like this table, floating in space. Exactly. The ground applies the force on the tree, and since the ground is not a part of the system, the force is an external one.
