Initially, mass one (2.10 kg) has a velocity of 5.90 m/s and mass two (2.80 kg) is at rest. After they collide, mass one emerges at an angle theta = 33.0 degrees. What is the speed of mass one after the collision if the collision is completely elastic? (Note, there are actually two possible answers two this problem, choose the solution which has m1 going as fast as possible.)
What is the angle phi, between mass two's velocity and the initial velocity of mass one? (Give your answer as a positive number in degrees.)
What is the final speed of mass two after the collision?
To solve this problem, we can utilize the principle of conservation of momentum and conservation of kinetic energy for an elastic collision.
1. First, let's compute the initial total momentum before the collision:
Initial momentum (p_initial) = (mass one * velocity one) + (mass two * velocity two)
p_initial = (2.10 kg * 5.90 m/s) + (2.80 kg * 0 m/s)
p_initial = 12.39 kg·m/s
2. Next, let's compute the initial total kinetic energy before the collision:
Initial kinetic energy (KE_initial) = 0.5 * mass one * (velocity one)^2 + 0.5 * mass two * (velocity two)^2
KE_initial = 0.5 * 2.10 kg * (5.90 m/s)^2 + 0.5 * 2.80 kg * (0 m/s)^2
KE_initial = 40.431 J
3. Since the collision is completely elastic, both momentum and kinetic energy are conserved after the collision.
4. Let's denote the final velocity of mass one as v_final and the final velocity of mass two as u_final.
5. Applying the conservation of momentum:
Final momentum (p_final) = (mass one * v_final) + (mass two * u_final)
Since mass two is initially at rest, the equation simplifies to:
p_final = mass one * v_final
Using the principle of conservation of momentum, p_initial = p_final:
12.39 kg·m/s = 2.10 kg * v_final
v_final = 5.90 m/s
6. Applying the conservation of kinetic energy:
Final kinetic energy (KE_final) = 0.5 * mass one * (v_final)^2 + 0.5 * mass two * (u_final)^2
Since mass two is initially at rest, the equation simplifies to:
KE_final = 0.5 * mass one * (v_final)^2
Using the principle of conservation of kinetic energy, KE_initial = KE_final:
40.431 J = 0.5 * 2.10 kg * (5.90 m/s)^2
Solving for v_final:
v_final ≈ 5.90 m/s
Therefore, the speed of mass one after the collision is approximately 5.90 m/s.
7. To find the angle phi between mass two's velocity and the initial velocity of mass one, we need to use trigonometry. Let's consider the right-angled triangle formed by the vectors of velocity one, velocity two (which is zero), and velocity final.
Theta (angle between velocity one and velocity final) is given as 33.0 degrees.
The angle phi can be calculated using the equation: phi = 90 degrees - Theta
phi = 90 degrees - 33.0 degrees
phi ≈ 57.0 degrees (rounded to one decimal place)
Therefore, the angle phi is approximately 57.0 degrees.
8. Since mass two was initially at rest and there was no external force acting on it after the collision, its final speed will be the same as the initial speed of mass one.
Hence, the final speed of mass two after the collision is 5.90 m/s.