Collisions can be elastic or inelastic. Learn about what's conserved and not conserved during elastic and inelastic collisions.
Log in Juan Rojas 8 years agoPosted 8 years ago. Direct link to Juan Rojas's post “In Exercice 2b, solution:...” In Exercice 2b, solution: I do not understand how " we should recognize that the kinetic energy which must be dissipated in the block is now four times higher". Can someone explain this further? • (13 votes) Rodrigo Campos 8 years agoPosted 8 years ago. Direct link to Rodrigo Campos's post “I believe he meant two ti...” I believe he meant two times higher. The kinetic energy carried by the bullet in the previous problem was mv²/2, while now is (m/2).(2v)²/2 = mv². Since all the kinetic energy is transferred to the block, it dissipates twice as much energy (13 votes) White 7 years agoPosted 7 years ago. Direct link to White's post “>Object A collides with a...” >Object A collides with an equal mass object B. Objects have equal but oppositely directed velocity. Does that mean If I use a 2-ball Newton's Cradle (sealed in vacuum, not friction of any kind, no energy loss). If I release the two balls at the same height and at the same time, they'll bounce of each other forever? • (7 votes) Charles LaCour 7 years agoPosted 7 years ago. Direct link to Charles LaCour's post “If there is no energy los...” If there is no energy loss then yes they will bounce for ever. (8 votes) Victoria 8 years agoPosted 8 years ago. Direct link to Victoria's post “In Exercise 1b, what was ...” In Exercise 1b, what was the formula used to solve for the problem? I can see that the KE formula wouldn't really work, since we don't know the final velocity of the racket. So which formula was it that it used to solve for the final velocity of the shuttle? • (8 votes) Asuruturato 7 years agoPosted 7 years ago. Direct link to Asuruturato's post “It was formula where the ...” It was formula where the book has derived from combination of conservation of energies and momentum, which is vBf=(2mAmA+mB)vAi+(mB−mAmA+mB)vBi Because that vBi was there in the example, it simply cancels out and our equation will get in that form. (0 votes) david romine 7 years agoPosted 7 years ago. Direct link to david romine's post “"A light object collides ...” "A light object collides with a much heavier target which is at rest. • (4 votes) Maybe the question is a little confusing?? Not sure since I dont see the whole thing. How about if the quesiton was rephrased: "A ball moving with velocity v strikes a wall and bounces off with velocity -v" (This means that the collision was perfectly elastic) Any thoughts?.. (4 votes) Anglachel 8 years agoPosted 8 years ago. Direct link to Anglachel's post “Alright, I dont get it. I...” Alright, I dont get it. In excersise 1b... In my way: Where is this wrong, I dont get... • (2 votes) Aibek Zhylkaidarov 8 years agoPosted 8 years ago. Direct link to Aibek Zhylkaidarov's post “I we can't use only kinet...” I we can't use only kinetic formula to find Vfs, because we DON'T know final velocity of racket after collision, So vfr is NOT equal to 0 (7 votes) DJ 7 years agoPosted 7 years ago. Direct link to DJ's post “In second case of elastic...” In second case of elastic collision that is: • (3 votes) V_Keyd 7 years agoPosted 7 years ago. Direct link to V_Keyd's post “Vaf = Vbi and Vbf = Vai a...” Vaf = Vbi and Vbf = Vai are the correct expressions. For ex: According to your assumption of B having an initial velocity along the negative axis, you will find that Vaf = - Vbi. Had I assumed B to be initially moving in a positive direction, ie., initial velocity of object B = +Vbi and A in a negative direction, - Vai, i would have found the final result to: Vaf = +Vbf. (3 votes) david romine 7 years agoPosted 7 years ago. Direct link to david romine's post “"Some kinetic energy gets...” "Some kinetic energy gets transformed into heat, sound, and used to deform the block. However, momentum must still be conserved." Does this mean that the loss in kinetic energy is small or that the loss in kinetic energy has a small affect on the momentum? With the momentum and kinetic energy coming from the projectile and some of the kinetic energy being transferred to the surroundings after impact then 1/2mv^2 decreases? • (3 votes) Teacher Mackenzie (UK) 7 years agoPosted 7 years ago. Direct link to Teacher Mackenzie (UK)'s post “Its a difficult concept t...” Its a difficult concept to work on. But you need to keep the following in mind... Somehow: Momentum is always conserved AS momentum. (Does not convert to energy) Kinetic energy is rarely conserved. Only in perfectly elastic collisions. BUT total energy is always conserved in whatever form. Here is a useful thing to remember about inelastic collisions In a perfectly inelastic collision (ie when the objects 'stick together' or coalesce, the MAXIMUM amount of KE is lost. Not necessarily ALL of it. Hope that helps (2 votes) Pradyun Mohamed 5 years agoPosted 5 years ago. Direct link to Pradyun Mohamed's post “in an inelastic collision...” in an inelastic collision, if kinetic energy is lost, does that mean the body is moving slower?, and if that is the case, how is momentum conserved? • (3 votes) Andrew M 5 years agoPosted 5 years ago. Direct link to Andrew M's post “momentum is a vector, KE ...” momentum is a vector, KE is scalar. Bodies moving in opposite directions have offsetting momentum, but the KE does not offset. (1 vote) redcloud4k13 4 years agoPosted 4 years ago. Direct link to redcloud4k13's post “how momentum is conserved...” how momentum is conserved if KE energy is lost? • (3 votes) Green 4 years agoPosted 4 years ago. Direct link to Green's post “If KE is lost then PE is ...” If KE is lost then PE is gained. Momentum doesn't have dependence on KE. Remember: p=mv. (1 vote) sushantsharma1994 6 years agoPosted 6 years ago. Direct link to sushantsharma1994's post “Could it be possible for ...” Could it be possible for coefficient of restitution to be 0 and it is not a prefectly inelastic collision? • (3 votes) daniel.rebrov 10 months agoPosted 10 months ago. Direct link to daniel.rebrov's post “if the coefficient is 0, ...” if the coefficient is 0, then Vf must also be 0. If the final speed after the collision is 0, then it is by definition a perfectly inelastic collision. (1 vote)Want to join the conversation?
The light object bounces off the target, maintaining the same speed but with opposite direction. The heavy target remains at rest." How can the heaver target remain at rest after having a force applied from the collision?
vfr = velocity, final, racket
vir = velocity, initial, racket
mr = mass, racket
ms = mass, shuttle
..
(mr×vir²)/2+(ms×vis²)/2 = (mr×vfr²)/2+(ms×vfs²)/2
(5×0)/2+(100×20²)/2 = (5×vfr²)/2+(100×0²)/2
(100×20²)/2 = (5×vfr²)/2
100*20²=5×vfr
(100*20²)/5=*vfr²
8000=vfr²
63,24=vfr
Object A collides with an equal mass object B. Objects have equal but oppositely directed velocity.
It state that both mass of objects are equal and velocity of both object should be in opposite direction.
so if we take initial velocity of object A is positive and initial velocity of object B is negative.
so after putting those value in formula of final velocity. We get
Vaf = - Vbi and Vbf = Vai instead of Vaf = Vbi and Vbf = Vai
Please check and reply Thanks.
David Sir you done a great job as usual..
In the formula shown in the article, instead of using minus sign to denote a velocity along the negative axis, simply Vi and Vf are used to get a general relation (one that will give the correct answer according to the initial velocity directions.)
The given expression for final velocities work for both cases.
I like this phrase because it reminds us that the priority here is conservation of momentum. Once that has been honoured and calculated, then we can determine the resulting kinetic energy.
