Ball 1, with a mass of 154 g and traveling at 12.6 m/s , collides head on with b

Question

Ball 1, with a mass of 154 g and traveling at 12.6 m/s , collides head on with ball 2, which has a mass of 329 g and is initially at rest. Assume that ball 1 is traveling to the right before the collision and that the collision is one dimensional.

a) What is the final velocity of ball 1 if the collision is perfectly elastic?

b) What is the final velocity of ball 2 if the collision is perfectly elastic?

C) What are the final velocities of the combination if the balls stick together?

d) How much KE was lost when the balls stuck together?

Give your answer as a positive value since the question asks how much was lost.

Explanation / Answer

m1 = 154 kg                   m2 = 329 kg

speeds before collision

u1 = 12.6 m/s                   u2 = 0 m/s

speeds after collision

v1 = ?                         v2 = ?

initial momentum before collision

Pi = m1*u1 + m2*u2

after collision final momentum

Pf = m1*v1 + m2*v2

from moentum conservation

total momentum is conserved

Pf = Pi

m1*u1 + m2*u2 = m1*v1 + m2*v2 .....(1)

from energy conservation

total kinetic energy before collision = total kinetic energy after collision

KEi = 0.5*m1*u1^2 + 0.5*m2*u2^2

KEf =   0.5*m1*v1^2 + 0.5*m2*v2^2

KEi = KEf

0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2 .....(2)

solving 1&2

we get

v1 = [ ((m1-m2)*u1) + (2*m2*u2) ] /(m1+m2)

a)

v1 = ( ((154-329)*12.6) + (2*329*0) ) /(154+329)

v1 = -4.56 m/s

b)

v2 = ( ((m2-m1)*u2) + (2*m1*u1) ) /(m1+m2)

v2 = ( ((329-154)*0) + (2*154*12.6) ) /(154+329)

v2 = 8.03 m/s

++++++++++++++

c)

for perfect inelastic

V1 = V2 = V

Pf = (m1+m2)*V

Pf = Pi

(154+329)*V = (154*12.6)

V = 4.02 m/s

-------------

KE loss = 0.5*m1*u1^2 - 0.5*(m1+m2)*v^2

KE loss = (0.5*0.154*12.6^2)-(0.5*(0.154+0.329)*4.02^2) = 8.32 J

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