Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
In a two-body collision in the center-of-mass reference frame the momenta of the particles are equal and opposite to one an- other both before and after the collision. Is the line of relative motion necessarily the same after collision as before? Under what conditions would the magnitudes of the velocities of the bodies increase? decrease? remain the same as a result of the collision? In a two-body collision in the center-of-mass reference frame the momenta of the particles are equal and opposite to one an- other both before and after the collision. Is the line of relative motion necessarily the same after collision as before? Under what conditions would the magnitudes of the velocities of the bodies increase? decrease? remain the same as a result of the collision?
Yes, the line of relative motions necessarily the same after collision as before.In accordance to the law of conservation of momentum, the total momentum of the particle having mass m1 and the particle having mass m2 before the collision equals their total momentum after the collision. The changes in momentum of the two objects have equal magnitudes and opposite signs, a necessary consequence of the law of conservation of momentum. Therefore the line of relative motions necessarily the same after collision as before.In an elastic collision, the final velocity v1f of the body having mass m1 is,v1f = (m1 – m2/ m1 + m2) v1i + (2m2/ m1 + m2) v2iand the final velocity v2f of the body having mass m2 is,v2f = (2m1/ m1 + m2) v1i + (m2 – m1/ m1 + m2) v2iHere, v1i is the initial velocity of the body having mass m1 and v2i is the initial velocity of the body having mass m2.As the target particle (m2) is at rest, thus, v2i = 0.Thus, v1f = (m1 – m2/ m1 + m2) v1i and v2f = (2m1/ m1 + m2) v1iTherefore the magnitude of the velocities of the bodies will depend upon the mass of the two bodies and initial velocities of the two bodies by the above two equations.As, m2 » m1, thus,v1f ≈ - v1i + 2 v2i and v2f = v2iWhen the massive particle is moving slowly or at rest, thenv1f ≈ - v1i and v2f = 0As, m1 >> m2, thus,v1f = v1i and v2f ≈ 2v1i - v2iTherefore, as, m2 >> m1 or m1 >> m2, then the final velocity of the body will be equal to the initial velocity of the body.
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -