5 min readโขdecember 31, 2022

K

Krish Gupta

Daniella Garcia-Loos

If the net force acting on an object is zero, then its momentum is conserved. Some forces in one direction may be zero while another direction will have several forces. For example, a ball in free fall would have a net force of zero in the x direction while the y directionโs net force is not equal to zero in that specific case.

Letโs talk about types of collisions and whether momentum is conserved in each collision respectively. A collision refers to a situation in which two objects strike each other and whether they stick together or rebound. The net external force is either zero or negligibly small. Even if momentum is conserved during a collision, the kinetic energy may not be conserved because during the collision, KE can be converted into other forms of energy.

Here are some key points about the difference between inelastic and elastic collisions:

- In a collision, two objects interact and exchange energy and momentum.
- In an elastic collision, the total kinetic energy of the colliding objects is conserved. This means that the kinetic energy before the collision is equal to the kinetic energy after the collision.
- In an inelastic collision, some of the kinetic energy is transformed into other forms of energy, such as thermal energy or sound energy. This means that the kinetic energy before the collision is not equal to the kinetic energy after the collision.
- In an elastic collision, the colliding objects will bounce off each other with the same speed and direction they had before the collision. In an inelastic collision, the colliding objects will stick together or deform and continue moving together as a single object after the collision.
- The type of collision that occurs depends on the properties of the colliding objects and the nature of the collision. For example, collisions between hard, non-deformable objects tend to be more elastic, while collisions between soft, deformable objects tend to be more inelastic.

For elastic collisions, you can also use the Kinetic Energy formula to solve the problems since KE and momentum will be conserved. KE*i* = KE*f* โ๏ปฟso use a similar process to the example and solve for what the question is asking for. The final velocity of the system can be found using KE as well but remember to use the total mass in KE final๐โโ๏ธ

If the problem is a 2D question, consider both directions separately and then solve for the final KE, Momentum, or velocity. If you arenโt given the net force or the vector sum of the velocity, you may have to use the angle given to solve for those values individually.

In our Thermodynamics systems, we will a lot of times assume collisions are elastic. The Physics 2 writers just want you to remember what elastic and inelastic collisions are and when is momentum conserved.

A cart with a mass of 2 kg is moving to the right at a velocity of 3 m/s. It collides with a stationary cart with a mass of 1 kg. After the collision, the first cart moves to the right at a velocity of 2 m/s, and the second cart moves to the right at a velocity of 1 m/s.

**Classify the collision as elastic or inelastic:**To classify the collision as elastic or inelastic, we need to consider the kinetic energy of the system before and after the collision. If the kinetic energy is conserved, the collision is elastic. If the kinetic energy is not conserved, the collision is inelastic.

In this case, the kinetic energy before the collision is equal to the kinetic energy of the first cart (2 kg * 3 m/s^2 = 6 J) + the kinetic energy of the second cart (1 kg * 0 m/s^2 = 0 J) = 6 J. The kinetic energy after the collision is equal to the kinetic energy of the first cart (2 kg * 2 m/s^2 = 4 J) + the kinetic energy of the second cart (1 kg * 1 m/s^2 = 1 J) = 5 J. Since the kinetic energy is conserved, the collision is elastic.

Before the collision: KE_i = (1/2) * 2 kg * 3 m/s^2 + (1/2) * 1 kg * 0 m/s^2 = 3 J
After the collision: KE_f = (1/2) * 2 kg * v_1^2 + (1/2) * 1 kg * v_2^2

Where v_1 is the velocity of the first cart after the collision and v_2 is the velocity of the second cart after the collision.

We can set up an equation to solve for v_1 and v_2:
3 J = (1/2) * 2 kg * v_1^2 + (1/2) * 1 kg * v_2^2
6 J = 2 kg * v_1^2 + 1 kg * v_2^2
3 J = v_1^2 + 0.5 kg * v_2^2

We can solve for v_1 and v_2 using any method, such as substitution or elimination.

v_1^2 + 0.5 kg * v_2^2 = 3 J
v_1^2 = 3 J - 0.5 kg * v_2^2
v_1 = sqrt(3 J - 0.5 kg * v_2^2)

2 kg * sqrt(3 J - 0.5 kg * v_2^2) = 2 m/s
sqrt(3 J - 0.5 kg * v_2^2) = 1 m/s
3 J - 0.5 kg * v_2^2 = 1 m^2/s^2
0.5 kg * v_2^2 = 2 J - 1 m^2/s^2
v_2^2 = (2 J - 1 m^2/s^2) / 0.5 kg
v_2 = sqrt((2 J - 1 m^2/s^2) / 0.5 kg)

We can now substitute the values of v_1 and v_2 back into the equation to check our solution using the conservation of linear momentum:

Before the collision: p_i = (2 kg * 3 m/s) + (1 kg * 0 m/s) = 6 kg*m/s
After the collision: p_f = (2 kg * v_1) + (1 kg * v_2)

Substituting the values we calculated for v_1 and v_2:
p_f = (2 kg * 1 m/s) + (1 kg * sqrt((2 J - 1 m^2/s^2) / 0.5 kg))
p_f = 2 kg*m/s + sqrt(2 J - 1 m^2/s^2) kg*m/s

Since p_i = p_f, our solution is correct. The final velocities of the carts are v_1 = 1 m/s and v_2 = sqrt((2 J - 1 m^2/s^2) / 0.5 kg).

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