1) A cart of mass 2.0kg is moving to the right along a smooth, horizontal rack at 3.0m/s. A Hooke's Law spring, of force constant 1200N/m and normal length 0.25m, is attached to its front. It collides "head-on" with a 4.0kg cart, initially at rest. What is the velocity of each cart at minimum spearation? [I don't have any idea of how to solve this problem. In all the problems I have done so far, it's asking for conditions before or after the collision, but in this case, it's asking something DURING the collision, which I am not sure how to handle...] 2) A bullet of mass 4.0g, moving horizontally with a velocity of 500m/s, strikes a wooden block of mass 2.0kg, initially at rest on a rough, horizontal surface. The bullet passes through the block in a negligible time interval, emerging with a velocity of 100m/s and casuing the block to slide 40cm along the surface before coming to rest. Find the maximum kinetic energy of the block and find the decrease in kinetic erergy of the bullet. Explain why the decrease in Ek of the bullet and the maximum Ek of the block are not equal. What happened to this difference in energy? [I have calculated that the maximum kinetic energy of the block = 0.64J and the decrease in kinetic erergy of the bullet=480J. But I can't explain why they aren't equal and what happened to this difference in energy...isn't the collision elastic?] Thank you for replying! Please Register or Log in to view the hidden image!
#1 Can you guess what the relationship between the 2 velocities would be at minimum separation? #2 What makes you think the collision is elastic?
1) Is there a relationship between them? I don't know... Please Register or Log in to view the hidden image! 2) Becuase for completely inelastic collisions, the 2 objects stick together and have the same final velocity. And in this case the 2 obejcts are not at all sticking together, so I guess it's elastic. Please Register or Log in to view the hidden image!
Let's call the 2kg cart "A" and the other one "B", with A moving to the right towards B. • If A is moving faster than B, then it's catching up with B, and the distance between them decreases. • If A is moving slower than B, then A is falling behind B, and the distance between them increases. The minimum distance is when A stops catching up with B and starts falling behind B. Can you guess what this means A's speed is, compared with B's? Careful :m: Some collisions get close to being perfectly elastic, while others get close to being perfectly inelastic, but most are somewhere in between. You can't conclude that a collision is elastic just because the final velocities are different.
I think you are suggesting that at the moment of minimum separation, the 2 carts have the same velocity. :m: But is it also possible that B is moving faster than A at the moment of minimum separation? (A started to fall behind...right?)
I am. Not at the moment of minimum separation. Only after. If B were moving faster than A, you could trace back a small time interval, like a millisecond, and find that A and B were closer at some point in the past, and your point wouldn't be the minimum separation. You can easily check all this for yourself if you've learned some differential calculus. The alternative is to do horrible calculations just to get the same result.
Oh, I see! It makes more sense now Please Register or Log in to view the hidden image! But is momentum conserved at any time? I know that the total momentum BEFORE the collision equals to the total momentum AFTER the collision, but how about the total momentum DURING the collision? Is it equal to the total momentum before the collision? Because as I know, for example, kinetic energy is not conserved DURING the collision
Momentum (like energy) is always conserved. The difference is that energy can change from one form to another, but there's only one "form" of momentum.
Another Sir Isaac Newton Contest Question: 3) A 1.0x10^3 kg plane is trying to make a forced landing on the deck of a 2.0x10^3 kg barge at rest on the surface of a calm sea. The only frictional force to consider is between the plane's wheels and the deck; this braking force is constant and is equal to one-quarter of the plane's weight. What must the minimum length of the barge be for the plane to stop safely on deck, if the plane touches down just at the rear end of the deck with a velocity of 5.0x10^1 m/s toward the front of the barge? Ff=2450N If the plane lands on the ground, the displacement can be calculated to be = 510.204m, but I know that the concept of momentum has to be used somewhere. I don't know where to use it...can somebody help me, please? The answer provided is 3.4x10^2 m, by the way. Please Register or Log in to view the hidden image!
Remember what you noticed earlier, about what happens when a moving object collides with a stationary one? Can they both be at rest afterwards?
They can't both be at rest. The problem here is that I don't know where to use the law of conservation of momentum. Please Register or Log in to view the hidden image! Please help me! Where did the collision occur, by the way?
4) A 0.25kg tennis ball is placed right on top of a 1kg volleyball and dropped. Both balls hit the ground at a speed of 3 m/s simultaneously. Find the (upward) velocity of the tennis ball right after it bounces up from the volleyball. Assume elastic collisions. (HINT: the tennis ball will move faster than 3m/s) Another tough question.......... Please Register or Log in to view the hidden image! I would appreciate if someone can help me! Please Register or Log in to view the hidden image!
But you assumed that the barge would be at rest after the plane landed... You can treat the interaction between the plane and barge like a collision (collisions don't have to be head-on). Remember that the reaction to the friction force will cause the barge to accelerate. Also, what does the final velocity of the plane have to be? Does it have to be zero?
I'd assume an elastic collision between the volleyball and the Earth, followed by an elastic collision between the volleyball and the tennis ball.
Hello, But it says that the tennis ball is placed right on top of the volleyball. How can the tennis ball and the volleyball collide if they are both moving at 3 m/s at impact? Will the volleyball rebound and leave the ground at a velocity 3 m/s[upward] (same as impact velocity with the ground)? If so, how can I justify this idea?
The final velocity of the plane should be zero because it stopped at the end, right? And the barge will also be at rest then... The velocity of the plane just before the interaction is 50m/s The velocity of the barge just before the interaction is 0m/s The velocity of the plane just after the interaction is 0m/s So the velocity of the barge just after the interaction can be found, it must be in the same direction as the orignal direction of the plane. :m: However, friction between the wheels and the deck exists, is momentum still convserved in this system?
You'll get instantaneous momentum transfer. It'll act like any other collision. Yes. If you want to justify it, you'll have to consider an elastic collision between the volleyball and the Earth. You should be able to show that, since the Earth's mass is so large compared with the volleyball, the volleyball will rebound at essentially the same speed as it hit the ground.
While the volleyball rebounds and move upward at 3m/s, wouldn't the tennis ball also be moving upward at 3m/s? It really seems to me that, at any point in time, the 2 balls are moving at the same velocity......then how can they collide after all? For elastic collision, the total kinetic energy is conserved. Part of the kinetic energy of the ball becomes the kinetic energy of the earth, and the earth has a large mass, so even a tiny speed would result in large kinetic energy. Wouldn't the kinetic energy of the ball after the collision be less because the original kinetic energy is now separated into two parts!?
I read over this part again, and the idea that "at minimum separation, the two carts have the same velocity" is making so much sense. Thanks for your nice explanation! Please Register or Log in to view the hidden image!
The tennisball doesn't hit the ground - only the volleyball does, so only the volleyball has its momentum inverted. When the volleyball hits the ground, it 'wants' to move up at 3 m/s, and the tennis ball 'wants' to continue falling at 3/s. Result: Elastic collision between the volleyball and the tennis ball. Here's a problem: Consider a small mass (m) with initial velocity u, and a larger mass (M) initially at rest. m hits M in an elastic collision. a) What's the final velocity, v, of the smaller mass, in terms of m, M, and u? b) What can you say about v (compared with u) when M is much larger than m? If you can solve this (it's a routine elastic collision, so it shouldn't be too difficult), you'll see for yourself that v is very close to -u if m is the 1 kg volleyball and M is the 6 sextillion tonne Earth.
