Elastic collision of two particles: The first particle has a mass of 0.1 kg and

Question

Elastic collision of two particles: The first particle has a mass of 0.1 kg and has a velocity of 1 m/sec. The second particle has the same mass and has the same velocity (in the opposite direction) a) What is the velocity of the two particles after the collision (assume that the incoming particle moves to the right)? b) Compute the velocity of the center of mass, before and after the collision. c) Assume that the duration of the collision is 0.01 seconds. What will be the impulse on each particle? d) What will the kinetic energy of each particle before and after the collision?

Explanation / Answer

2.

a)

m1 = 0.1 kg      m2 = 0.1 kg

v1i = 1 m/s      v2i = - 1 m/s

v1f = ?      v2f = ?

using conservation of momentum

m1 v1i + m2 v2i = m1 v1f + m2 v2f

(0.1) (1) + (0.1) (-1) = (0.1) v1f + (0.1) v2f

v1f = - v2f                                      Eq-1

using conservation of kinetic energy

m1 v21i + m2 v22i = m1 v21f + m2 v22f

(0.1) (1)2 + (0.1) (-1)2 = (0.1) v21f + (0.1) v21f

V1f = - 1 m/s

and V2f = 1 m/s

b)

Vbefore = m1 v1i + m2 v2i /(m1 + m2) = (0.1) (1) + (0.1) (-1) /(0.2) = 0 m/s

Vafter = m1 v1f + m2 v2f /(m1 + m2) = (0.1) (-1) + (0.1) (1) /(0.2) = 0 m/s

c)

Change in momentum = m (V2f - V2i) = (0.1) (1 - (-1)) = 0.2

t = time of collision = 0.01 sec

Impulse = change in momentum / time = 0.2 /0.01 = 20 Ns

d)

KEbefore = (0.5) (0.1) (1)2 = 0.05 J

KEafter = (0.5) (0.1) (-1)2 = 0.05 J

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