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Possibly your confusion arises from not considering that KE is a scalar, whilst momentum is a vector. Yes, there is of course a connection between KE and momentum: $K = p^2/2m$ (for non-relativistic bodies). Thus, for two equal-mass particles heading directly towards each other at equal speeds $v$, their velocities are $\pm {\bf v}$ and their momenta $\pm ... 4 I'm 27 and since I was about 15 I had the same doubt you do. Only a couple of years ago I realized why momentum is always conserved in a collision, whereas the same is not enforced for energy. (They must have told me this at some point -- or points --, but I guess sometimes I just don't pay much attention) First of all, let's make it empirically clear that ... 4 Is there no relation between the momentum and kinetic energy? Are they not linked with each other? No, they're not. Conservation of momentum only says that $$m_1 v_{1_-} + m_2 v_{2_-} = m_1 v_{1_+} + m_2 v_{2_+}$$ I used subscripts 1 and 2 denote the particles involved in the collision and the subscripts - and + denote the pre- and post-collision ... 2 The only thing that the device "knows" when it is hit, is the force with which it gets hit, and the duration of that hit. Transfer of momentum$m\Delta v = F\Delta t$. So what matters is the momentum of the hammer's head - or more specifically, the momentum that you are able to transfer. Ultimately it comes down to giving the most momentum to the head of the ... 2 If the kinetic energy of both the ball decreases, how can their velocities be equal?? The front one ( B ),from the beginning, had low KE; if it decreases during the deformation, how can its velocity be equal to the velocity of the rear ball ( A )? In order to get a clear picture, let's consider the extreme case when the velocity of B = 0 Let's ... 2 You're right, it is fairly obvious. Conservation of linear momentum applies at all times! If an object is irradiated by alpha particles, each with mass$\sim 4m_u$and travelling with velocity$v$, and if it absorbs$N$of these particles in a time$t$. Then the force exerted on the object is the rate of change of momentum. $$F = \frac{4Nm_u v}{t}$$ Of ... 1 Books For fundamentals I prefer books over papers, because they are typically more thorough and a little bit more 'slow' in the introduction of concepts. There are many books that will cover, some of, the topics that you mention. I will mention below the 3 books that where most useful to me in the past. 1) An excellent resource for a theoretical foundation ... 1 As some other materials in addition of those definitely helpful references proposed by Michiel, I found the following very useful resources: a full review of many concepts in droplet dynamics has been presented in this book: Ashgriz, N. Handbook of Atomization and Sprays. Vol. 11. Springer New York, 2010. for droplet collision: Rein, Martin. ... 1 The kinetic energy of the center of masses of the two colliding bodies may increase if their internal structure changes, i.e. if at least one of the bodies were in an excited state, and the conservation laws of energy and momentum allow the exceeding energy to transform into kinetic energy. 1 Consider the relative velocity of one body with respect to another. That's all. So, relative velocity of approach would mean the velocity of body 1 coming towards body 2, signified by$v_1-v_2$And the relative velocity of separation would mean the velocity of body 1 going away from body 2, signified by$u_1-u_2$as seen by body 2, in both the cases Yes, ... 1 Suppose total kinetic energy were conserved (i.e. if you can ignore potential energy changes), then because it's a nonlinear fuunction of the velocities, the sum of the kinetic energies of the particles that make up a composite body does not equal the kinetic energy of the center the center of mass of the object. In practice, what this means is that you ... 1 It came from the wall. If a ball hits a wall at constant velocity, then it's not going to "slow down and eventually come to rest". It's going to bounce back, and in fact, if the collision is elastic, it won't even necessarily slow down. The force on the ball from the wall serves to change the direction of the velocity (and possibly decrease its magnitude ... 1 Let's say that the ball strikes the floor with the linear momentum$-p_1$(you see I take the direction of the momentum in consideration). Some part of the momentum is lost (some energy is lost to the floor) s.t. the ball rises again, but at the floor level it has a linear momentum$+p_2$. So, what the floor "gets" in this scenario? It GETS a linear momentum ... 1 Think of it this way: Ball A is moving at 10 m/s Ball B is moving at 3 m/s Both balls have the same mass Here we have Ball A collides with Ball B, transferring energy. During this transfer (ignore the deformation for now, since that doesn't seem to be an issue at the moment, so we've got an elastic collision), consider what happens in the exchange of ... 1 by conservation of linear momentum the center of mass of the system will keep moving, thus the two ball are together they will keep moving in the same direction than the faster ball, at the same speed than the center of mass, that is, slower than the initial speed of the faster ball. Assuming a totally elastic collision, after the balls push back each other, ... 1 Let$S$and$S'$be the two inertial frame and$S'$moving with a constant velocity v w.r.t.$S$frame. Now a force$F$acting on the particle at point$A$and displace it to the point$B$. If the position x-coordinates of A point and B point in$S'$frame are$(x_1',x_2')$and in$S$frame are$(x_1,x_2)$then at any time$t$,$x_1=x_1'+vt\$ and ...