A 3kg ball moving at 8 m/s strikes a 2kg ball at rest,if the collision is elastic,what is the speed of the lighter ball if the heavier ball moves at 2m/s in the opposite direction

Answer:

Speed of lighter ball is 4 m/s.

Explanation:

Applying the principle of conservation of linear momentum,

momentum before collision = momentum after collision.

[tex]m_{1}[/tex][tex]u_{1}[/tex] + [tex]m_{2}[/tex] [tex]u_{2}[/tex] = [tex]m_{1}[/tex][tex]v_{1}[/tex] - [tex]m_{2}[/tex][tex]v_{2}[/tex]

[tex]m_{1}[/tex] = 3 kg, [tex]u_{1}[/tex] = 8 m/s, [tex]m_{2}[/tex] = 2 kg, [tex]u_{2}[/tex] = 0 m/s ( since it is at rest), [tex]v_{1}[/tex] = 2 m/s, [tex]v_{2}[/tex] = ?

(3 x 8) + (2 x 0) = (8 x 2) - (2 x [tex]v_{2}[/tex])

24 + 0 = 16 - 2[tex]v_{2}[/tex]

2[tex]v_{2}[/tex] = 16 - 24

2[tex]v_{2}[/tex] = -8

[tex]v_{2}[/tex] = [tex]\frac{-8}{2}[/tex]

   = -4 m/s

This implies that the light ball moves at the speed of 4 m/s in the opposite direction of the heavier ball after collision.

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o
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