A 0.10-kg cart traveling in the positive x direction at 10.0 m/s collides with a

Question

A 0.10-kg cart traveling in the positive x direction at 10.0 m/s collides with a 0.30-kg cart at rest. The collision is elastic. What is the velocity of the 0.10-kg cart after the collision? I understand the answer to be -5 m/s. However, can someone aid in simplifying the principle of conservation of linear momentum to derive this answers? m1vf1 + m2vf2 = m1vi1 + m2vi2 vf: velocity final vi: velocity initial Thank you!!

Explanation / Answer

m1vf1 + m2vf2 = m1vi1 + m2vi2 0.10*vf1 + 0.30vf2 = 0.10*10 + 0.30*0 0.1vf1 + 0.30vf2 = 1 and convervation of energy ; m1vf1^2 + m2vf2^2 = m1vi1^2 + m2vi2 ^2 0.10*vf1^2 + 0.30vf2^2 = 0.10*10^2 + 0.30*0^2 0.1vf1^2 + 0.3vf2^2 = 10 simplify, 0.1vf1 + 0.30vf2 = 1 --> vf2 = (1 - 0.1vf1)/0.3 0.1*vf1^2 + 0.3{(1-0.1vf1)/0.3} ^2 = 10 0.1 vf1^2 + (1-0.1vf1^2)/0.3 = 10 solving gives, v = -5 m/s

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