# Help me with physics question pls!!!?

A ball with mass mA= 0.500 kg, moving to +x direction with a speed of 4.50 m/s has a head-on collision with a 1.000 kg ball, which is initially at rest. Considering that the collision is perfectly elastic, find the speed and direction of each ball after the collision.

Thank you

Relevance

Velocity first ball vA2 = -1.5 m/s

Velocity second ball vB2 = 3 m/s

"collision is perfectly elastic" means that both conservation of energy and conservation of impulse applies.

Energy before and after

mA vA1^2 / 2 = mA vA2^2 / 2 + mB vB2^2 / 2

or

mA vA1^2 = mA vA2^2 + mB vB2^2

here

mA and mB = masses

vA1 = velocity ball A before

vA2 = velocity ball A after

vB1 = velocity ball B before (vB1=0)

vB2 = velocity ball B after

Impulse before and after

mA vA1 = mA vA2 + mB vB2

From equation of impulse

mA vA2 = mA vA1 - mB vB2

vA2 = (mA vA1 - mB vB2) / mA

square and substitute into equation for energy

mA vA1^2 = (mA vA1 - mB vB2)^2 / mA + mB vB2^2

mA^2 vA1^2 = (mA vA1 - mB vB2)^2 + mAmB vB2^2

0 = - 2 mA mB vA1 vB2 + mB^2 vB2^2 + mAmB vB2^2

(mB^2 + mAmB) vB2^2 - 2 mA mB vA1 vB2 = 0

((mB^2 + mAmB) vB2 - 2 mA mB vA1) vB2 = 0

(mB^2 + mAmB) vB2 - 2 mA mB vA1 = 0

vB2 = 2 mA mB vA1 / (mB^2 + mAmB)

vB2 = 2 * 0.5 * 1 * 4.5 / (1^2 + 0.5*1)

vB2 = 3m/s

vA2 = (0.5 * 4.5 - 1 * 3) / 0.5 = -1.5 m/s

Check

Energy before/after

0.5 * 4.5^2 = 0.5 * (-1.5)^2 + 1 * 3^2

10.125 = 10.125 ....... OK

Impulse

0.5 * 4.5 = 0.5 (-1.5) + 1 * 3

2.25 = 2.25

Edit: 