Question

--Physics

--PROBLEM #2

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2) Consider a head-on collision of the particle of mass M, moving with the velocity V and having kinetic energy E=MV2/2, with another particle of mass M1, staying at rest. Find the energy Ej transferred to the second particle in the course of the collision.

Answer 1

Before Collision M V | ✓-o After collision mi Vi from Conservation of Energy 1 mu? = 1 mv 2 + 1 mint mv²= mv2 + min? | 0 Conservation of momentin MV to = mu't mini I mv= mv'tmi vill [V= v'+ mv (from2) on putting in eo x moi vi +2m1v") = mint ve Il mi 2 vit mir I mil m( vt mil y ) = mvi? A minh m - Milyt av v= mi v + Zv' = mvi? + miri2 my + kimi ? vi + 2 miv'vi m = myla tmv vinmi v = 20' vici-mi = 21 m ma tamir'v, = mira IV (1 m) = 21 (m_M) v = 2V IVI = 2 ( m m.)

Energy transferred to the second particle in course of collesen Change in KE of 1st particle (mass m) KE lost by ME KE gamed by my LOKE = 1 mivie -of (Initially it was prest) - (for massimi) M-mi 9 mm 4m (4 m2 x ) - 2mm. KE trem fere dy 12m 1 .m.) 20:27 we may also determined it by calculating the Energy lost by Massimo - 1 mv 2 - 1 mv 12 we know ve v't mi v M-mi 2MI V = v'+ 2m1 vi = villt ami ) (M.-+2m) - VY Man m-m M+MI M-M +2ml M-mi M-MI N mtml m-

Now KElost mt-1 mvi Emriz KEI No to Imv 21 m2 * L rn ( mm ) - Love mm mviz (Matmi) ?- (m-mi)? (m-mi) 2 2/(metyn, + M-mi (M+mix+mi)) L mot (M-mi) ? Imv2 4 m Levey (2010) 2m (Mamy w I which samme as 6

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