Question

Masses my and m2, traveling with velocities +v; and -u respectively have a head on collision. After the collision, mass mi is at rest while mass me moves with velocity +us (same direction as +v). a) Draw a sketch of both situations, labeling the information you've been given (3 points) b) Without assuming anything about Energy, prove that m2 = m (7 points) c) Suppose vy is such that the collision is completely INELASTIC. What is m2 in terms of my only? (4 points) d) Suppose vy is such that the collision is completely ELASTIC. What is m2 in terms of mı only? (6 points)

Answer 1

solution - mass of object be "m, & m2" & its velocity "vi &- vi? Befse collision}. - mass of object be "m, & m» & its velocity "o & tuf (same direction as tvil & Affer Collision mi m2 Ti ve 7 At ** mi mzur At rest. Before During head collision After collision. on collision, how of conservation 6 momentum must » Initial momentum - Final momentum mvi - m₂ Vi = mxo + Me Of m, ui – m₂ Vi= m₂ Of m2 (vit vf) = m vi - mivi (vitus This is sollution of past b.

② Completely together inelastic collision after collision & means these particle stick is maximum. lost of kinetic energy 7 Coefficient of restitutione =0 K. Eafter collision is zen > The velocity of chi tm) 18 zero after. collision (ufo) ut=0 m (mitm₂) mi vi u haw of conservation of momentum Initial momentum Final momentum mivi - ma vi = (mitm) XO - 11III This is result for O. d Completely Elastic means law of conservation of momentum as well as of kinetic Chergy will hold. my m, vi mi meni ? Rest.

m2 By law of conservation of momentom. mui- m₂ Vi = m xo + mauf » 04 (mm) Vi -- 0 2 By law of conservation of momentom kinetic energy- Ima 2 + + m₂ Vi² = Im xo + 1 m2 lg 2 N temt ma) Vi= 1 muz? semi-mener 22 (m + m₂) ve ² = 1 m2 x ma vie (mitm₂) - (mm2) 2. I mitm2) = Don mz uuu me mit M2 ² = m 2+ M₂² – gm, m₂ 3 m, m₂ = m 2 [m2 = my This is relation between m & m2 for. too perfectly elastic collisian. mza