A block of mass m = 4.4 kg,moving on frictionless surface with a speed v_i = 9.2m/s, Makes a sudden perfectly elastic collision with a second block of mass Mass shown in the figure.The second block is originally at rest.Just after the collision,the 4.4-kg block recoils with a speed of v_f=2.5m/s. What is the mass M of the second block? A)7.7 kg B)0 kg C)0...kg D)0.. kg E)0.kg

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Quizplus · Accepted Answer

In a perfectly elastic collision, momentum is conserved. Let's denote the mass of the second block as M. Before the collision: Momentum = m*v_i (since the second block is at rest) After the collision: Momentum = m*v_f + M*v2 where v2 is the velocity of the second block after the collision. Since the collision is perfectly elastic, kinetic energy is also conserved: 1/2*m*v_i^2 = 1/2*m*v_f^2 + 1/2*M*v2^2 Substituting the given values, we can solve for M: 4.4*9.2 = 4.4*2.5 + M*v2 1/2*4.4*9.2^2 = 1/2*4.4*2.5^2 + 1/2*M*v2^2 Solving these equations, we get: M = 7.7 kg

A Block of Mass M = 4