# 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

### 1 Answer

- SlowfingerLv 62 months agoFavourite answer
Answer:

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

discard solution 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:

"prettier" formula layout added