A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.540 m/s. The block has a head-on elastic collision with a 40.0 gram block that is initially at rest. Since the collision is head-on, all velocities lie along the same line, both before and after the collision. (a) What is the speed of the 20.0 gram block after the collision

  0.180 m/s

Explanation:

Solving the equations for conservation of momentum and energy for an elastic collision gives ...

  v₁' = ((m₁ -m2)v₁ +2m₂v₂)/(m₁ +m₂) . . . . v₁' is the velocity of m₁ after collision

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Here, we have (m₁, m₂, v₁, v₂) = (20 g, 40 g, 0.540 m/s, 0 m/s).

Substituting these values in to the equation for v₁', we have ...

  v₁' = ((20 -40)(0.540) +2(40)(0))/(20 +40) = (-20/60)(0.540)

  v₁' = -0.180 . . . m/s

The speed of the 20 g block after the collision is 0.180 m/s to the left.

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explanation:

newton's first law of motion