A One-Dimensional Inelastic Collision Part A Block 1, of mass m 3.90 kg, moves a

Question

A One-Dimensional Inelastic Collision Part A Block 1, of mass m 3.90 kg, moves along a frictionless air track with speed vi15.0 m/s. It collides with block 2, of mass m2 57.0 kg , which was initially at rest. The blocks stick together after the collision. (Figure 1) Find the magnitude pi of the total initial momentum of the two-block system Express your answer numerically Pi m/s Submit Hints My Answers Give Up Review Part Part B Find vf, the magnitude of the final velocity of the two-block system Express your answer numerically Figure 1 Uf m/s Before collision: Submit Hints My Answers Give Up Review Part Part C What is the change K = Kinal-Kinitial in the two-block system's kinetic energy due to the collision? After collision: Express your answer numerically in joules.

Explanation / Answer

conservation of momentum
initial momentum = final momentum
   m1u1 +m2u2 = m1v1 +m2v2
   m1u1 + m2u2 = (m1+m2)v

part A)       initial momentum = 3.90 *15 = 58.5 kg m/s
           3.90 *15 + 57*0 = (3.9 +57)v

part b)           final velocity v = m1u1 + m2u2 /(m1+m2)
                   = 3.90 *15 + 57*0 /(3.9 +57)  
                   = 0.960 m/s
part c)  
    initial kinetic energy 1/2 m1u1^2 = 1/2 *3.9*225=438.75 J
   final kinetic energy of the system = 1/2 (m1+m2) v^2
                   = 1/2*(60.9)*0.960^2
                   = 28.06 J

   change in kinetic energy = 28.06 - 438.75 = -410.69 J

RELATION BETWEEN MOMENTUM AND KINETIC ENERGY
   K.E = P^2/2M
   P = sqrt(2*K.E*M)
BOTH HAVING SMAE K.E then the ratio of their momentum is
   p_c/ P_b = sqrt(m_c/m_b)
  

       = sqrt(4.4*10^-2/0.144)
       = 0.5527

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