Mastering Elastic Collisions: AP Physics Problems with Solutions

[mrbetty16]

I need help with some of these problems. Any kind of help will be most appreciated

1. Speed amplifier. In Fig. 10-3, block 1 of mass m1 slides along an x-axis on a frictionless floor with a speed of v1i = 2.80 m/s. Then it undergoes a one-dimensional elastic collision with stationary block 2 of mass m2 = 0.200m1. Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass m3 = 0.200m2

2. Speed deamplifier. In Fig. 10-5, block 1 of mass m1 slides along an x-axis on a frictionless floor with a speed of 2.40 m/s. Then it undergoes a one-dimensional elastic collision with stationary block 2 of mass m2 = 2.70m1. Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass m3 = 2.70m2.

3. A block of mass m1 = 2.2 kg slides along a frictionless table with a speed of 8 m/s. Directly in front of it, and moving in the same direction, is a block of mass m2 = 5.0 kg moving at 2.8 m/s. A massless spring with spring constant k = 1100 N/m is attached to the near side of m2, as shown in Fig. 10-35. When the blocks collide, what is the maximum compression of the spring? (Hint: At the moment of maximum compression of the spring, the two blocks move as one. Find the velocity by noting that the collision is completely inelastic to this point.)

 

[dynamicsolo]

I need help with some of these problems. Any kind of help will be most appreciated

1. Speed amplifier. In Fig. 10-3, block 1 of mass m1 slides along an x-axis on a frictionless floor with a speed of v1i = 2.80 m/s. Then it undergoes a one-dimensional elastic collision with stationary block 2 of mass m2 = 0.200m1. Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass m3 = 0.200m2

2. Speed deamplifier. In Fig. 10-5, block 1 of mass m1 slides along an x-axis on a frictionless floor with a speed of 2.40 m/s. Then it undergoes a one-dimensional elastic collision with stationary block 2 of mass m2 = 2.70m1. Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass m3 = 2.70m2.

The first two involve the same basic analysis. Have you had the equation relating the velocities of the masses in a one-dimensional elastic collision? Combining conservation of linear momentum with conservation of kinetic energy leads to the conclusion that the relative velocity of the two masses after the collison is the negative of the relative velocity before the collision, or

v1 - v2 = -( v1' - v2' ) [ or, in its easier-to-remember form, v1 + v1' = v2 + v2' .

If you use this together with the equation for linear momentum conservation
(don't forget to use the correct signs of the masses' velocities in these equations), it is relatively easy to solve for the velocities. In each problem, you will have two collisions to solve successively.

3. A block of mass m1 = 2.2 kg slides along a frictionless table with a speed of 8 m/s. Directly in front of it, and moving in the same direction, is a block of mass m2 = 5.0 kg moving at 2.8 m/s. A massless spring with spring constant k = 1100 N/m is attached to the near side of m2, as shown in Fig. 10-35. When the blocks collide, what is the maximum compression of the spring? (Hint: At the moment of maximum compression of the spring, the two blocks move as one. Find the velocity by noting that the collision is completely inelastic to this point.)

You can solve this in a stationary reference frame with some effort, but the center-of-mass frame is really the way to go. Find the velocities of the two blocks in this frame before the collision (keep in mind that in this reference frame, the total momentum of the blocks is zero!) and calculate their kinetic energies. The spring is maximally compressed at the moment when the velocities of both blocks in this frame is zero (corresponding to having the same velocity in the stationary frame). That means that all the kinetic energy of the blocks has gone into the potential energy of the spring. Since you have the (ideal) spring constant, it is straightforward to find the maximum compression of the spring. [With a little trouble, it can be shown that, although the observed kinetic energies differ between the two reference frames, the result for the spring compression is the same -- as it must be, since the displacement from equilibrium of the spring should not depend on its relative velocity to the observer (at least classically).]

 

[ogb p]

I would be happy to provide assistance with these problems. Elastic collisions can be challenging, but with a few key principles and some practice, they can be mastered.

For the first problem, it is important to remember that in an elastic collision, both momentum and kinetic energy are conserved. This means that the total momentum before the collision must equal the total momentum after the collision, and the total kinetic energy before the collision must equal the total kinetic energy after the collision.

To solve this problem, you can start by writing down the initial and final momentum equations for each collision. In the first collision, the initial momentum of block 1 is m1v1i, and the final momentum is m1v1f. For block 2, the initial momentum is 0, since it is stationary, and the final momentum is m2v2f. Using the conservation of momentum, you can set these two equations equal to each other and solve for v1f and v2f. Then, using the conservation of kinetic energy, you can set the initial kinetic energy of block 1 (1/2m1v1i^2) equal to the final kinetic energy of both blocks (1/2m1v1f^2 + 1/2m2v2f^2) and solve for v1f again. This will give you the final velocity of block 1 after the first collision.

You can then repeat this process for the second collision between blocks 2 and 3. Once you have the final velocity of block 2 after this collision, you can use the conservation of momentum and kinetic energy once again to solve for the final velocity of block 3.

For the second problem, the process is the same, but you will have to be careful with the signs of the velocities since block 2 is now moving in the opposite direction.

For the third problem, the key is to recognize that the collision between the two blocks is completely inelastic until the moment of maximum compression of the spring. This means that the two blocks will move as one until that point. Using this information, you can set up an equation for the conservation of momentum for the two blocks together, and solve for their final velocity. This will give you the velocity at the moment of maximum compression of the spring. Then, using the spring equation F = -kx, where F is the force exerted by the spring, k is the spring