M1 = 150g = 0.150 kg.
M2 = 100g = 0.10kg.
M1*V1 + M2*u = M1*4 + M2*4.
0.150*0 + 0.10*u = 0.15*4 + 0.10*4.
0.1u = 1.0, u = 10 m/s.
KE1 = 0.5M*u^2 = 0.5*0.1 * 10^2 = 5 Joules = Total kinetic energy before collision.
KE2 = 0.5M1*V1^2 + 0.5*M2*V2.
KE2 = 0.5*0.15*4^2 + 0.5*0.10*4^2 = 2 Joules = Total KE after collision.
KE Lost = KE1-KE2.
A particle A of mass 150g lies at rest on a smooth horizontal surface. A second particle B of mass 100g is projected along the surface with a speed of u m/s and collides directly with A. On collision the masses coalesce and moves on with a speed of 4 m.s . find the value of u and the loss in the kinetic energy of the system during impact.
If m1u1 + m2u2 = m1v1 + m2v2
Then does 150/1000u1 + 100/1000u2 = (150 + 100)4?
Then how would you proceed? I feel like I'm missing a trick but I can't get the answer.
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