Blocks A (mass 2.50 kg) and B (mass 10.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 5.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let +x be the direction of the initial motion of A.
1.) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time.
Find the maximum energy.
2.) Find the velocity of A.
3.) Find the velocity of B.
4.) Find the velocity of each block after they have moved apart.
Find the velocity of A.
5.) Find the velocity of B.
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy. Let's go step by step:
1.) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time.
To find the maximum energy stored in the spring bumpers, we need to consider that at the maximum compression or extension of the springs, the blocks momentarily stop and reverse their directions. At that point, all of their initial kinetic energy is converted into potential energy stored in the springs.
Initially, block A is moving towards block B with a velocity of 5.00 m/s. Block B is at rest. Since there is no external force acting on the system, the total momentum before the collision is conserved.
Initial momentum, p_initial = (mass_A x velocity_A) + (mass_B x velocity_B)
Since velocity_B is zero initially, p_initial = (mass_A x velocity_A)
When both blocks reach their maximum compression or extension, their velocities become zero (since they momentarily stop).
Final momentum, p_final = (mass_A x 0) + (mass_B x 0) = 0
Using the conservation of momentum, we have: p_initial = p_final
(mass_A x velocity_A) = 0
Solving for velocity_A at maximum compression/extension, we get velocity_A = 0.
Now, to find the maximum energy stored in the spring bumpers, we can use the conservation of kinetic energy.
Initial kinetic energy, KE_initial = (1/2 x mass_A x velocity_A^2) + (1/2 x mass_B x velocity_B^2)
At maximum compression/extension, the velocity of both blocks is zero, so KE_final = 0.
Using the conservation of kinetic energy, we have: KE_initial = KE_final
(1/2 x mass_A x velocity_A^2) + (1/2 x mass_B x velocity_B^2) = 0
Substituting the values obtained earlier, we have: (1/2 x 2.50 kg x 0^2) + (1/2 x 10.00 kg x 0^2) = 0
Therefore, the maximum energy stored in the spring bumpers is 0 Joules, and the velocity of each block at maximum compression/extension is 0 m/s.
2.) Find the velocity of A.
We already determined that the velocity of block A at maximum compression/extension is 0 m/s.
3.) Find the velocity of B.
Since there is no external force acting on the system, the total momentum is conserved. The momentum of block A before the collision is equal to the momentum of block B after the collision.
Initial momentum of block A = (mass_A x initial velocity_A)
Final momentum of block B = (mass_B x final velocity_B)
Using the conservation of momentum, we have: (mass_A x initial velocity_A) = (mass_B x final velocity_B)
(2.50 kg x 5.00 m/s) = (10.00 kg x final velocity_B)
Solving for final velocity_B, we get: final velocity_B = (2.50 kg x 5.00 m/s) / 10.00 kg
Therefore, the velocity of block B after the collision is 1.25 m/s.
4.) Find the velocity of each block after they have moved apart.
Since the collision is elastic and the blocks are equipped with ideal spring bumpers, the total mechanical energy (kinetic + potential energy) of the system is conserved.
Initially, the kinetic energy of the system is given by: KE_initial = (1/2 x mass_A x initial velocity_A^2) + (1/2 x mass_B x initial velocity_B^2)
KE_initial = (1/2 x 2.50 kg x 5.00 m/s^2) + (1/2 x 10.00 kg x 0 m/s)^2
At maximum compression/extension (when the velocities are momentarily zero), the system only has potential energy stored in the springs:
Potential energy at maximum compression/extension, PE_max = KE_initial
Finally, at maximum compression/extension, the potential energy of the system (stored in the springs) is converted back to kinetic energy as the blocks move apart.
Using the conservation of mechanical energy, we have: KE_initial = KE_final + PE_max
(1/2 x 2.50 kg x velocity_A^2) + (1/2 x 10.00 kg x velocity_B^2) = 0 + 0
Substituting the known values, we have: (1/2 x 2.50 kg x velocity_A^2) + (1/2 x 10.00 kg x 1.25 m/s)^2 = 0
Solving for velocity_A after they have moved apart, we can rearrange the equation to isolate velocity_A:
velocity_A^2 = [-(1/2 x 10.00 kg x 1.25 m/s)^2] / (1/2 x 2.50 kg)
Taking the square root of both sides, we get: velocity_A = √[-(1/2 x 10.00 kg x 1.25 m/s)^2] / (1/2 x 2.50 kg)
Simplifying this expression, we find: velocity_A = -√(1/2 x 10.00 kg x 1.25 m/s)^2 / (1/2 x 2.50 kg)
Therefore, the velocity of block A after they have moved apart is approximately -2.50 m/s (negative because it is in the opposite direction of the initial motion).
5.) Find the velocity of B.
Since we already determined that the velocity of block A after they have moved apart is -2.50 m/s, we can use the conservation of momentum to find the velocity of block B.
Using the conservation of momentum, we have: (mass_A x velocity_A) + (mass_B x velocity_B) = 0
(2.50 kg x -2.50 m/s) + (10.00 kg x velocity_B) = 0
Solving for velocity_B, we get: velocity_B = (2.50 kg x -2.50 m/s) / 10.00 kg
Therefore, the velocity of block B after they have moved apart is approximately -0.625 m/s (negative because it is in the opposite direction of the initial motion).