Collisions, Fractional Momentum

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In summary, the momentum imparted to the ball is 1.0 Ns and the velocity imparted is 2.0 m/s. The equations for momentum and energy conservation are used to calculate the final velocities for the two masses after the collision. The fractional momentum transfers, Q1 and Q2, are calculated for two cases and it is found that in Case 2, the transfers are larger.
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1. A force F=2.0N is applied for 0.5sec to a ball of mass m=0.5kg at rest on a smooth surface (no friction). Determine the momentum P and the velocity v imparted to the ball due to the impulse from the applied force F.

2. A ball of mass m1=0.5kg moving with a velocity v01=2.0m/sec moving in a straight line makes an elastic collision with a ball of mass m2 of mass 0.5kg at rest (v02=0.0m/sec) on a smooth surface (no friction).

a) Write down the equations for momentum conservation and energy conversation for this system using vf1 and vf2 as the final velocities for the two masses after the collision.

b) The expressions for vf1 and vf2 were derived in class as

vf1=v01(m1-m2/m1+m2)
vf2=v01(2m1/m1+m2)

Calculate the fractional momentum transfers, Q1 and Q2, where:
Q1=(Po1-Pf1/Po1) and P01=m1v01 and Pf1=m1vf1

Q2=(Po2-Pf2/Po1) and P02=m2v02 and Pf2=m2vf2

Calculate Q1 and Q2 for the following two cases
Case 1: m1=0.5kg, m2=0.5kg
Case 2: m1=1.9kg, m2=0.5kg

Use v01=2.0m/sec and v02=0.0m/sec for both cases.

c) In which of the Cases, 1 and 2, are the fractional momentum transfers larger?

I've attempted this homework but got stuck. For my momentum in question 1. I got a value of 4 and a velocity of 2. I'm lost in everything else. thank you for your help!

Answer:1. The momentum imparted to the ball is P = F x t = 2.0 N x 0.5 s = 1.0 Ns. The velocity imparted to the ball is v = P/m = 1.0 Ns / 0.5 kg = 2.0 m/s. a) Momentum conservation equation: m1v01 + m2v02 = m1vf1 + m2vf2 Energy conservation equation: ½ m1v01² + ½ m2v02² = ½ m1vf1² + ½ m2vf2² b) For Case 1, Q1 = (1.0 Ns - 0.0 Ns) / 1.0 Ns = 1.0 and Q2 = (0.0 Ns - 1.0 Ns) / 0.0 Ns = -1.0. For Case 2, Q1 = (3.8 Ns - 0.6 Ns) / 3.8 Ns = 0.84 and Q2 = (0.0 Ns - 1.8 Ns) / 0.0 Ns = -1.8. c) In Case 2, the fractional momentum transfers are larger.

I would like to provide a response to the content.

Firstly, in question 1, we can calculate the momentum P by using the equation P=mv, where m is the mass of the ball and v is the velocity imparted. Therefore, P= (0.5kg)(2m/s)= 1kgm/s. Similarly, we can calculate the velocity v by using the equation v= FΔt/m, where F is the applied force, Δt is the time interval and m is the mass of the ball. Therefore, v= (2N)(0.5s)/(0.5kg)= 4m/s.

In question 2, an elastic collision is taking place between two balls of equal mass. In this case, the equations for momentum conservation and energy conservation can be written as:

Momentum conservation: m1v01 + m2v02 = m1vf1 + m2vf2

Energy conservation: ½ m1v01^2 + ½ m2v02^2 = ½ m1vf1^2 + ½ m2vf2^2

a) Using vf1 and vf2 as the final velocities for the two masses after the collision, we can solve for the final velocities by substituting the given values into the above equations.

b) The expressions for vf1 and vf2 were derived in class as:

vf1= v01(m1-m2)/ (m1+m2)

vf2= 2v01m1/ (m1+m2)

To calculate the fractional momentum transfers, we can use the formula Q= (Po-Pf)/Po, where Po is the initial momentum and Pf is the final momentum.

For Case 1:
Po1= m1v01= (0.5kg)(2m/s)= 1kgm/s
Pf1= m1vf1= (0.5kg)(2m/s)= 1kgm/s
Therefore, Q1= (1-1)/1= 0

Po2= m2v02= (0.5kg)(0m/s)= 0
Pf2= m2vf2= (0.5kg)(2m/s)= 1kgm/s
Therefore, Q2= (0-1)/0= undefined

For Case 2:

1. What is a collision in physics?

A collision in physics is defined as an event where two or more objects come into contact with each other, resulting in a change in their motion and energy. Collisions can be elastic or inelastic, depending on whether or not kinetic energy is conserved.

2. How is momentum related to collisions?

Momentum is a key factor in collisions, as it is a measure of an object's motion. In a closed system, the total momentum before a collision is equal to the total momentum after the collision. This is known as the law of conservation of momentum.

3. What is fractional momentum?

Fractional momentum, also known as reduced mass, is a concept used in collisions involving two or more objects with different masses. It is a measure of the effective mass of a system and is calculated by taking the product of the masses and dividing it by their sum.

4. How does fractional momentum affect the outcome of a collision?

Fractional momentum plays a crucial role in determining the outcome of a collision. In a two-object collision, the object with a higher fractional momentum will experience a smaller change in velocity compared to the object with a lower fractional momentum.

5. What are some real-life applications of collisions and fractional momentum?

Collisions and fractional momentum have many real-life applications, such as in car crashes, sports, and particle accelerators. Understanding these concepts allows us to predict and analyze the outcomes of these events and make improvements to safety measures or technological advancements.

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