Question

n procedure 3: assume cart 1 has mass 220 g and initial velocity +v10 = 4.38 m/s, and cart 2 has mass 819 g. Assume the track is frictionless, and after the "elastic" collision cart 1 is moving at v1 = -2.429 m/s and cart 2 is moving at v2 = 1.786 m/s. a) What percentage of the original energy was lost in this collision? % b) What percentage of the original linear momentum was "lost" due to external forces? (Remember: linear momentum is a vector!) p = %

Answer 1

a)
m1 = 0.22kg

m2 = 0.819 kg

v1o = 4.38 m/s
v1 = -2.429 m/s
v2 = 1.786 m/s

let v2o is the initial velocity of the second cart.

intial momentum = final momentum

m1*v1o + m2*v2o = m1*v1 + m2*v2

==> v2o = (m1*v1+m2*v2 - m1*v1o)/m2

= (-0.22*2.429 + 0.819*1.786 - 0.22*4.38)/0.819

= -0.043 m/s (velosity of cart 2 before collsion)

Ki = 0.5*m1*v1o^2 + 0.5*m2*v2o^2

= 0.5*0.22*4.38^2 + 0.5*0.819*0.043^2

= 2.11 J

Kf = 0.5*m1*v1^2 + 0.5*m2*v2^2

= 0.5*0.22*2.429^2 + 0.5*0.819*0.1.786^2

= 1.955 J

% of loss of energy = (Ki-kf)/Ki *100

= 7.35 %

b) there is no loss of momentum in any collsion.