Need help with this physics momentum/collision problem please. Two billiard ball

Question

Need help with this physics momentum/collision problem please.

Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is moving upward along the y-axis at 2.0 m/s, and ball B is moving to the right along the x-axis with speed 3.7m/s. After the collision (assumed elastic), the second ball is moving along the positive y-axis (See figure). What is the speed of ball A after the collision? What is the speed of ball B after the collision? In what direction is ball A moving after the collision? What is the total momentum of the two balls after the collision? What is the total kinetic energy of the two balls after the collision?

Explanation / Answer

(a), (b)

For elastic collision,

1/2*m*vA^2+1/2*m*vb^2 =1/2*m*vA'^2+1/2*m*vb'^2

vA^2+vb^2 =vA'^2+vb'^2

(c) both ball will turn 90 degress to its orinal direction, A going to +x and B going +y direction.

vA'^2 = 3.7^ +2^2 -vb^2

= 17.69 - vb^2

Now as we know,

m*va +m*vb = m*vA' + m*vb'

vA' = 5.7 - vb'

Hence

(5.7 - vb')^2 = 17.69 - vb^2

32.49 -11.4 vb +vb^2 = 17.69 - vb^2

2*vb^2 +11.4 vb -14.8 = 0

vb = [-11.4 +_ root ( 130+118.4) ]/4

vb = 1.09 m/s

Hence,

va = 5.07 -1.09 =3.17 m/s

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---