Final answer:
There may be a mistake in the given options or the problem statement. The calculated value for v_2f is approximately 3.2 m/s based on the given values.
The correct option is noy given.
Step-by-step explanation:
To solve this problem, we can use the conservation of linear momentum along the x-axis. The initial linear momentum of the system is equal to the final linear momentum. The formula for linear momentum p is p = m . v, where m is mass, and v is velocity.
Initially, only the first disk is moving, so the initial linear momentum p_initial is given by:
p_initial = m .v
After the collision, the two disks move off with final velocities v_1f and v_2f at angles (θ _1) and θ_2 with respect to the x-axis. The final linear momentum (p_final) is the sum of the linear momenta of the two disks:
p_final = m . v_{1f} . cos(θ_1) + m . v_{2f} . cos(tθ_2)
Since the second disk starts from rest v_{2i} = 0, its initial linear momentum p_2i is zero.
Now, set the initial linear momentum equal to the final linear momentum and solve for v_{2f}:
m . v = m . v_{1f} . cos(θ_1) + m . v_{2f} . cos(θ_2)
Cancel out m from both sides:
v = v_{1f} . cos(θ_1) + v_{2f} . cos(θ_2)
Now, plug in the given values:
10 m/s = 8 m/s . cos(30^\circ) + v_2f . cos(θ_2)
Evaluate the trigonometric functions:
10 = 8 . {3^1/2 / {2} + v_{2f} . cos(θ_2)
Solve for v_{2f}:
v_2f = {10 - 8 . 3^1/2 / 2 cos(θ_2)
Given that θ_2 is below the x-axis, cosθ_2 is negative. Now, calculate v_2f:
v_2f = {10 - 8 . {{3}^1/2} / {2}} / {cos(θ_2)}
v_2f ≈ {10 - 8 . {{3}^1/2} / {2}} / {cos(150^\circ)}
v_2f ≈ {10 - 8 . {{3}^1/2} / {2}} / {{{3}^1/2} / {2}}
v_2f ≈ {10 + 4{3}^1/2 / {3}^1/2
v_2f ≈ {10{3}^1/2 + 12} / {3}
v_2f ≈ {10{3}^1/2 + 12}{3} x {3}^1/2 / {3}^1/2
v_2f ≈ {30 {3}^1/2 + 36} / {9}
v_2f ≈ {10{3}^1/2 + 12} / {3}
v_2f ≈ 3.154 m/s
Rounding to one decimal place, v_2f ≈ 3.2 m/s.
So, the correct answer is not provided among the options. There might be a mistake in the given answer choices or the problem statement.