A 0 .1 kg pinball is fired horizontally by a springwith force constant 40 N/m. If the spring is depressed 10 cm and the ball collides with a 0.3 kg ball elastically, what is the post collision velocity of the 0.3 kg ball? *

Explanation:

We can find out the velocity of pinball after being fired from the equation of conservation of mechanical energy

1/2 m v² = 1/2 k x² , m is mass of ball , v is velocity after firing , k is spring constant , x is depression in spring.

.1 x v² = 40 x .1 x .1

v = 2 m /s

ball of mass .1 , collides with velocity 2 m /s ,

from the equation of elastic collision

v₂ = (m₂ - m₁) u₁ / (m₂ + m₁) + 2m₁m₂u₂ / (m₂ + m₁)

m₁ ,u₁ are mass and velocity of first ball , m₂ , u₂ are mass and velocity of second mass , v₂ is velocity of second mass after collision.

u₁ = 2 , u₂ = 0 , m₁ = .1  m₂ = .3

v₂ = (.3 - .1 ) x 2/ (.3 + .1 )

= (.2 / .4 )x 2

= 1 m /s

So .3 kg ball will move with velocity of 1 m /s in the same direction as .1 m mass collided with it.

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1 the 4 is sig

c. magnitude and direction