An air track glider of mass m1= 0.200kg moving at 0.750m/s to the right collides

Question

An air track glider of mass m1= 0.200kg moving at 0.750m/s to the right collides with a glider of mass   m2= 0.400kg at rest. If m1 rebounds and moves to the left with a speed of 0.250m/s, what is the speed and direction of m2 after the collision? (Hint-- Momentum is a vector quantity, and direction is indicated by the sign of momentum). Show work.

Calculate the original kinetic energy of the system before the collision. Calculate the total kinetic energy after the collision. What happened to the lost energy?

Explanation / Answer

Masses m = 0.2 kg            M = 0.4 kg Initial velocity of mass m is u = 0.75 m / s Initial velocity of mass M is U = 0 Velocity of mass m after collision v = -0.25 m / s Velocity of mass M after collsion V = ? From law of conservation of momentum , mu + MU = mv + MV                                                                           mu = mv +MV From this V = ( mu - mv ) / M                     = m( u - v) / M                     = 0.2( 0.75 +0.25 ) / 0.4                    = 0.2 / 0.4                    = 0.5 m/ s (b). Kinetic energy before collision K = ( 1/ 2) mu 2 + ( 1/ 2) MU 2                                                          = 0.05625 J + 0                                                          = 0.05625 J (c). Kinetic energy after colliison K ' = ( 1/ 2) mv 2 + ( 1/ 2) MV 2                                                         = (6.25 x 10 -3 ) + 0.05                                                         = 0.05625 J                                                         = (6.25 x 10 -3 ) + 0.05                                                         = 0.05625 J (d). Loss in kinetic energy = K - K '                                        = 0

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.