A body weighing 2 kg moves at a speed of 3 m / s and catches up with another body weighing 3 kg

| June 18, 2021 | education

A body weighing 2 kg moves at a speed of 3 m / s and catches up with another body weighing 3 kg, moving at a speed of 1 m / s. Find the velocities of the bodies after the collision if: 1) the impact was inelastic; 2) the impact is elastic. The bodies move in one straight line, the blow is central.

1) In both the first and second cases, the law of conservation of momentum is fulfilled. Let m = 2 kg and M = 3 kg be the masses of the bodies, v = 3 m / s and u = 1 m / s – the velocities of the bodies.
2) In the case of an inelastic impact, the bodies stuck together into one with a mass m + M. Let the velocity of this body be V, then we have: mv + Mu = (m + M) V, V = (mv + Mu) / (m + M) = (2 kg * 3 m / s + 3 kg * 1 m / s) / (2 kg + 3 kg) = 1.8 m / s.
3) In the case of an elastic impact, the bodies will fly apart in different directions. We have: mv + Mu = – mv1 + Mv2. Here v1 and v2 are the velocities of the bodies after collision.
4) We also have in the second case the law of conservation of energy: mv ^ 2/2 + Mu ^ 2/2 = mv1 ^ 2/2 + Mv2 ^ 2/2.
From this equation and the equation of paragraph 3, we obtain the equation-consequence: v – v1 = u + v ^ 2, v – u -v1 = v ^ 2. Substituting this into the equation of point 3, we get: v1 = | (Mv – 2Mu – mv) / (m + M) | = 2.4 m / s. Then v2 = 0.4 m / s.
ANSWER: 1) V = 1.8 m / s; 2) v1 = 2.4 m / s, v2 = 0.4 m / s.



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