in apool game,a cue ball traveling at 0.75 m/s hits the stationary eight ball The eight ball moves with a velocity of 0.25 m/s at an angle of 37 deg reletive to the cue balls inital direction assuming that the collision is inelastic at what angle will the que be deflected and what will be its speed?

its a glancing collision, i believe that the question improperly says the collision is inelastic. It should be elastic. there fore the sum the y and x vectors. substitute. You should get velocity .57 m/s and 15.2 degrees

Whether the collision is elastic or inelastic, the total momentum of the two balls in the direction perpendicular to the cue ball's initial direction must be zero, since there was no original momentum in that direction. Therefore

MV1 sin 37 - MV2 sin A = 0
In addition,
M Vo = M cos 37 + M V2 cos A
since momentum in the forward direction is conserved.
The M's all cancel out
Vo = 0.75 m/s
V1 = 0.25 m/s is the eight-ball's final velocity.
"A" is the unknown angle of the cue ball, and V2 is its velocity. It will be a negative angle.
You have two equations in two unknowns to solve.

We will be glad to critique your work.

The velocity is 0.0000 m/s

To find the angle at which the cue ball will be deflected and its speed after the collision, we can apply the principle of conservation of momentum in the x and y directions.

Let's denote the mass of the cue ball as mcue, the mass of the eight ball as meight, the initial velocity of the cue ball as vcue_initial, the final velocity of the eight ball as veight_final, the angle of deflection of the cue ball as θ, and the final velocity of the cue ball as vcue_final.

Using conservation of momentum in the x-direction:

mcue * vcue_initial + 0 = mcue * vcue_final * cos(θ) + meight * veight_final * cos(37°)

Simplifying this equation, we get:

mcue * vcue_initial = mcue * vcue_final * cos(θ) + meight * veight_final * cos(37°) ----(Equation 1)

Using conservation of momentum in the y-direction:

0 = mcue * vcue_final * sin(θ) + meight * veight_final * sin(37°)

Simplifying this equation, we get:

0 = mcue * vcue_final * sin(θ) + meight * veight_final * sin(37°) ----(Equation 2)

Now, we have two equations (Equation 1 and Equation 2) and two unknowns (θ and vcue_final). We can solve these equations simultaneously to find the values.

First, solve Equation 1 for vcue_final:

vcue_final = (mcue * vcue_initial - meight * veight_final * cos(37°)) / (mcue * cos(θ))

Next, substitute this value of vcue_final into Equation 2:

0 = mcue * ((mcue * vcue_initial - meight * veight_final * cos(37°)) / (mcue * cos(θ))) * sin(θ) + meight * veight_final * sin(37°)

Simplifying this equation, we get:

0 = (mcue * vcue_initial - meight * veight_final * cos(37°)) * sin(θ) + meight * veight_final * sin(37°)

Now, we can solve this equation for θ:

θ = arctan((meight * veight_final * sin(37°)) / (mcue * vcue_initial - meight * veight_final * cos(37°)))

Substituting the known values: mcue = meight = 1 (as masses cancel out), vcue_initial = 0.75 m/s, veight_final = 0.25 m/s, and θ = 37°, we can calculate the angle of deflection (θ):

θ = arctan((1 * 0.25 * sin(37°)) / (0.75 - 1 * 0.25 * cos(37°)))

Calculating this equation, we find:

θ ≈ 44.26°

Therefore, the cue ball will be deflected at an angle of approximately 44.26°.

Now, substitute the value of θ and the known values into Equation 1 to find vcue_final:

vcue_final = (1 * 0.75 - 1 * 0.25 * cos(37°)) / (1 * cos(44.26°))

Calculating this equation, we find:

vcue_final ≈ 0.79 m/s

Therefore, the speed of the cue ball after the collision will be approximately 0.79 m/s.

the answer is 0