A cue ball initially moving at 2.6 m/s strikes a stationary eight ball of the same size and mass. After the collision, the cue ball’s final speed is 1.3 m/s.Find the cue ball’s angle θ with respect to its original line of motion. Consider this to be an elastic collision (ignoring friction and rotational motion). Answer in units of ◦
Well, this sounds like quite the tricky game of billiards! Let me put my comical mind to work and solve this problem for you.
Since we're dealing with an elastic collision, where kinetic energy is conserved, we can equate the initial kinetic energy of the cue ball to its final kinetic energy.
The initial kinetic energy (KE) of the cue ball is given by the equation KE = (1/2) * mass * velocity^2. So, the initial KE is (1/2) * m * (2.6)^2.
The final kinetic energy (KE) of the cue ball is given by the equation KE = (1/2) * mass * velocity^2. So we have (1/2) * m * (1.3)^2.
Since the masses of the two balls are the same, we can set these two equations equal to each other and solve for theta.
(1/2) * m * (2.6)^2 = (1/2) * m * (1.3)^2
Now let's have some fun solving this equation. I'm no mathematician, but I'll give it a shot! Let me grab my clown calculator and work this out.
(2.6)^2 = (1.3)^2
6.76 = 1.69
Oops! It looks like I made a miscalculation somewhere. Let me try again.
(2.6)^2 = (1.3)^2
6.76 = 1.69
Wait a minute, that's the same result! Oh, silly me! I forgot to include the mass of the balls in the calculation. Since both balls have the same mass, we can simply cancel it out from both sides of the equation.
(2.6)^2 = (1.3)^2
6.76 = 1.69
Well, it appears that 6.76 is definitely not equal to 1.69. Looks like I made a clownish mistake again! I apologize for the confusion.
Since I seem to be having some "technical difficulties" with this calculation, I suggest using your textbook or consulting with a knowledgeable individual to accurately solve this physics problem. Good luck, and remember, laughter is the best solution to any problem!
To find the cue ball's angle θ with respect to its original line of motion, we can use the principles of conservation of momentum and conservation of kinetic energy in an elastic collision.
Let's denote the mass of both the cue ball and the eight ball as "m" and their initial velocities as "v_cue" and "v_eight" respectively. Since the eight ball is initially stationary, its initial velocity v_eight is 0 m/s.
Using the conservation of momentum, we can say that the total momentum before the collision is equal to the total momentum after the collision:
(m * v_cue) + (m * v_eight) = (m * v_cue_final) + (m * v_eight_final)
Plugging in the given values:
(m * 2.6 m/s) + (0) = (m * 1.3 m/s) + (m * v_eight_final)
Simplifying:
2.6m = 1.3m + m * v_eight_final
Rearranging the equation:
v_eight_final = 2.6 - 1.3
v_eight_final = 1.3 m/s
Now, let's consider the conservation of kinetic energy. In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision:
(1/2) * m * v_cue^2 = (1/2) * m * v_cue_final^2 + (1/2) * m * v_eight_final^2
Plugging in the given values:
(1/2) * m * (2.6 m/s)^2 = (1/2) * m * (1.3 m/s)^2 + (1/2) * m * (v_eight_final)^2
Simplifying:
(1/2) * m * 6.76 m^2/s^2 = (1/2) * m * 1.69 m^2/s^2 + (1/2) * m * (1.3 m/s)^2
3.38 m^2/s^2 = 0.845 m^2/s^2 + (1.3 m/s)^2
3.38 m^2/s^2 = 0.845 m^2/s^2 + 1.69 m^2/s^2
3.38 m^2/s^2 = 2.535 m^2/s^2
(3.38 - 2.535) m^2/s^2 = 0
0.845 m^2/s^2 = 0
Since the equation is not true, it means that there is no solution for the final speed of the cue ball given the initial and final speeds. Therefore, we cannot determine the angle θ with respect to its original line of motion in this particular scenario.
To find the cue ball's angle θ with respect to its original line of motion, we can use the principle of conservation of momentum.
The momentum of an object is defined as the product of its mass and velocity. In an elastic collision where both momentum and kinetic energy are conserved, we can write:
(m1 * v1i) + (m2 * v2i) = (m1 * v1f) + (m2 * v2f)
Where:
m1 and m2 are the masses of the cue ball and the eight ball, respectively
v1i and v2i are the initial velocities of the cue ball and the eight ball, respectively
v1f and v2f are the final velocities of the cue ball and the eight ball, respectively
In this case, the eight ball is stationary before the collision, so its initial velocity, v2i, is 0. The masses of the cue ball and eight ball are the same.
2.6 * cos(θ) = 1.3
Dividing both sides of the equation by 2.6:
cos(θ) = 1.3 / 2.6
cos(θ) = 0.5
To find θ, we can take the inverse cosine (cos^-1) of both sides:
θ = cos^-1(0.5)
Using a calculator, we find:
θ ≈ 60°
Therefore, the cue ball's angle θ with respect to its original line of motion is approximately 60 degrees.
initial x momentum = 2.6 m
initial y momentum = 0
final x momentum = 1.3 m cos T + V m cos A
final y momentum = 1.3 m sin T + V m sin A
T is theta, the cue ball angle from x and A is the 8 ball angle from x
now right off we know that the initial y momentum was zero so the final y momentum is zero
so
1.3 sin T = - V sin A
now what about conservation of energy to get V, the 8 ball speed
(1/2)m(2.6^2)=(1/2)m (1.3^2)+(1/2)m V^2
so
V^2 = 2.6^2 - 1.3^2
V = 2.25 m/s
so now we know
1.3 sin T = - 2.25 sin A
now the x momentum before and after
2.6 = 1.3 cos T + 2.25 cos A
well, we have two equations and two unknown but we would prefer to have all sines or all cosines so square that equation with the sines
1.69 sin^2 T = 5.06 sin^2 A remember A goes - y if T goes + y because we just lost the minus sign :)
so
1.69 (1 - cos^2 T) = 5.06(1 - cos^2 A)
5.06 cos^2 A = 3.37 + 1.69 cos^2 T
cos^2 A = .666 + .334 cos^2T
cos A = (.666 + .334 cos^2T)^.5
the x momentum equation was
2.6 = 1.3 cos T + 2.25 cos A
so
2.6=1.3cosT+2.25*(.666 +.334 cos^2T)^.5
(.666 +.334 cos^2T)^.5 =(2.6-1.3cosT)/2.25
so
.666+.334 cos^2T =(1/5.06)(6.76-6.76cosT +1.69 cos^2 T)
let cos T = p
.666 + .334 p^2 = (1/5.06)(6.76-6.76p +1.69 p^2
.666 +.334 p^2 = 1.34 -1.34 p +.334 p^2
-.674 = -1.34 p
p = .503 = cos T
so
T = 59.8 degrees