Inelastic 2D Collison?

Question

Two balls of equal mass collide. One ball is stationary when it was hit, after the collision they make an angle of 48.9 and 31.1. Find the speed of each ball after the collision if the intial speed of ball1 is 2.25m/s?

Answer 1

Use momentum conservation, dude.

Horizontal(direction of initial motion of movin ball):

mu + MU [Initial momentum of system]= mv cos48.9 + MV cos 31.1 [Final momentum of system]

=> u + U = v cos 48.9 + V cos 31.1 (coz m=M)  ..............  1

Vertical(perpendicular to direction of motion of movin ball):

0 [Coz no initial motion vertically] = mv sin 48.9 - MV sin 31.1(coz they move in opp directions)

=> v sin 48.9 = V sin 31.1   .................2

Solve 1 n 2 to find v and V.

Answer 2

The conservation of momentum applies to this problem:

For the horizontal components:

M1V1 + M2V2 = M1(V3)(cos 48.9) + M(V4)(cos 31.1)

and for the vertical components:

0 = M1(V3)(sin 48.9) + M2(V4)(sin 31.1)

where

M1 = mass of the first ball

M2 = mass of the second ball

V1 = initial speed of the first ball = 2.25 m/sec.

V2 = initial speed of the second ball = 0 (at rest)

V3 = speed of the first ball after the collision

V4 = speed of the second ball after the collision.

NOTE that since M1 = M2, then the horizontal components equation simplifies to

2.25 = V3(cos 48.9) + V4(cos 31.1)

2.25 = 0.6574V3 + 0.8563V4

and the vertical components equation becomes

V3(sin 48.9) = V4(sin 31.1)

V3 = V4(sin 31.1/sin 48.9)

V3 = 0.6854V4 --- substituting this in the simplified horizontal components equation

2.25 = 0.6574(0.6854V4) + 0.8563V4

2.25 = 0.4506V4 + 0.8563V4

1.3069V4 = 2.25

V4 = 1.72 m/sec.

Solving for V3,

V3 = 0.6854(1.72)

V3 = 1.18 m/sec.

ANSWER:

After the collision, the speed of the first ball is

V3 = 1.18 m/sec

and the speed of the second ball is

V4 = 1.72 m/sec.