Question

Let two objects of equal mass m collide. Object 1 has initial velocity v, directed to the right, and object 2 is initially stationary.
A. If the collision is perfectly elastic, what are the final velocities v1 and v2 of objects 1 and 2?

Give the velocity v1 of object 1 followed by the velocity v2 of object 2, separated by a comma. Express each velocity in terms of v.

B. Now suppose that the collision is perfectly inelastic. What are the velocitiesv1 and v2 of the two objects after the collision?

Give the velocity v1 of object 1 followed by the velocity v2 of object 2, separated by a comma. Express each velocity in terms of v.

C. Now assume that the mass of object 1 is 2m, while the mass of object 2 remains m. If the collision is elastic, what are the final velocities

v1 and v2 of objects 1 and 2?

Give the velocity v1 of object 1 followed by the velocity v2 of object 2, separated by a comma. Express each velocity in terms of v.

D. Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision?

Give the velocity v1 of object 1 followed by the velocity v2 of object 2, separated by a comma. Express each velocity in terms of v.

Answer 1

Answer:

A: v1=0, v2=v; B: 0.5v; C: v/3, 4v/3; D: 0.25 V

Explanation:

A)

In this case, the collision can be described by two laws: Momentum conversations law and Energy conservation law. Note, that theoretically, the vehicles will be moving in the unknown directions after the collision, but let's assume that the final velocities are going to have the same direction with the original velocity. Then the two equations are going to be:

$mV=mV1+mV2\\mV^{2}/2=mV1^{2}/2 + mV2^{2}/2$

After some transformations, the following can be derived:

$\left \{ {{V=V1+V2} \atop {V^{2} =V1^{2}+V2^{2} }} \right.$

After we square the top equation and substitute the squares of velocities, with the V^2 from the equation below, we are going to get the following:

$2V1V2=0$, from where we can make a logical conclusion: V1=0, as it cannot pass by the second vehicle. In this case, V2=V.

B)

In this case, only Momentum conservation law can be used, as the energy will be lost during the collision. After the collision, objects will move as one object with the same velocity. To find this velocity, we can do the following:

$mV=mV1+mV2\\V1=V2=U\\mV=mU+mU\\U=V/2=0.5V$

C) This case, similarly to the case A, requires to find two velocities. We can use both conservation laws- Momentum and Energy conservation laws. In this case, the mathematics will look as:

$2mV=2mV1+mV2\\2mV^{2}/2=2mV1^{2}/2 + mV2^{2}/2\\\\2V=2V1+V2\\2V^{2}=2V1^{2}+V2^{2}$

Substituting V2 in the second equation, we are getting:

$2V^{2}=2V1^{2}+4V^{2}-8V1 V +4V1^{2} \\6V1^{2}-8V1 V+2V^{2}=0$

From where, by solving a square equation, we are getting two roots:

V1= V and V1=V/3. First answer shows us the initial conditions and the second results- velocity after the collision.

Using this, we can find, that V2=4V/3

So, the answers are: V1=V/3; V2=4V/3

D)

In this case, both objects after the collision will move as a single object with the same velocities. Only the momentum will be conserved in this scenario, so only one equation is required:

$mV=mV1+3mV1\\mV=4mV1\\V1=V/4=V2$