Question

A ball of mass 0.250 kg and a velocity of + 5.00 m/s collides head-on with a ball of mass 0.800 kg that is initially at rest. No external forces act on the balls. After the collision, the velocity of the ball which was originally at rest is + 2.38 m/s. What is the velocity of the 0.250 kg ball?

Answer 1

mass of ball 1= 0.25 kg

initial speed of ball 1= 5. m/s

mass ball 2= 0.8 kg

initial speed ball 2= 0 ( it was at rest)

final speed of ball 2= 2.38m/s

the formula is:

m1 . v 1 + m2 . v2 = m1 . v1f+ m2 . v2f

0.25 . 5 + 0.8 . 0 = o.25 . x + 0.8 . 2.38

-2.6288 m/s

Step-by-step explanation:

i got it right on the flvs

Answer 2

`v_1=-2.616\ m/s`

Step-by-step explanation:

Given that,

Mass of ball 1,
`m_1=0.25\ kg`

Initial speed of ball 1,
`u_1=5\ m/s`

Mass of ball 2,
`m_2=0.8\ kg`

Initial speed of ball 2,
`u_2=0` (at rest)

After the collision,

Final speed of ball 2,
`v_2=2.38\ m/s`

Let
`v_2` is the final speed of ball 1.

Initial momentum of the system is :

`p_i=m_1u_1+m_2u_2`

`p_i=0.25* 5+0`

`p_i=1.25\ m/s`

Final momentum of the system is :

`p_f=m_1v_1+m_2v_2`

`p_f=0.25* v_1+0.8* 2.38`

`p_f=0.25 v_1+1.904`

According the law of conservation of linear momentum :

initial momentum = final momentum

`1.25=0.25 v_1+1.904`

`v_1=-2.616\ m/s`

So, the final velocity of ball 1 is (-2.616)m/s.