Momentum and Impulse. find final velocity

In summary, using the formula m1v1i+m2v2i=(m1v1f+m2v2f), the final velocity of the 0.10-kg object after an elastic head-on collision with a 0.15-kg object initially at rest is -0.045 m/s.
  • #1
maxjohnfresh
1
0

Homework Statement


A 0.10-kg object moving initially with a velocity of 0.20 m/s eastward makes an elastic head-on collision with a 0.15-kg object initially at rest. What is the final velocity of the .10 kg object after the collision ?


Homework Equations



im not sure which equation to use :/

The Attempt at a Solution



i tried using m1v1+m2v2=(m1+m2) V
but it didnt come out right.

and i know the answer is -0.045 m/s
 
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  • #2
Hi maxjohnfresh, welcome to PF.
i tried using m1v1+m2v2=(m1+m2) V
This formula is true if both objects move together after collision.
So use the formula
m1v1i+m2v2i=(m1v1f+m2v2f) .
Apply the conservation of energy and solve for v1f and v2f.
 
  • #3
eastward.


I would suggest using the equations for conservation of momentum and conservation of kinetic energy to solve this problem.

First, we can calculate the initial momentum of the system using the equation p = mv, where p is momentum, m is mass, and v is velocity. In this case, the initial momentum of the system is (0.10 kg)(0.20 m/s) = 0.02 kg*m/s.

Next, we can use the conservation of momentum equation, p1 + p2 = p1' + p2', where p1 and p2 are the initial momenta of the two objects, and p1' and p2' are the final momenta after the collision. Since p2 is initially at rest, its momentum is 0, so the equation becomes p1 = p1'.

Substituting in the values, we get 0.02 kg*m/s = (0.10 kg)v1'. Solving for v1', we get v1' = 0.02 kg*m/s / 0.10 kg = 0.2 m/s eastward.

Finally, we can use the conservation of kinetic energy equation, KE1 + KE2 = KE1' + KE2', where KE is kinetic energy. Since the collision is elastic, kinetic energy is conserved, so the equation becomes KE1 = KE1'. Substituting in the values, we get (1/2)(0.10 kg)(0.20 m/s)^2 = (1/2)(0.10 kg)(v1')^2. Solving for v1', we get v1' = 0.2 m/s.

Therefore, the final velocity of the 0.10 kg object after the collision is 0.2 m/s eastward. This is the same as the initial velocity, indicating that the collision was perfectly elastic and the object did not lose any kinetic energy.
 

1. What is the difference between momentum and impulse?

Momentum is the measure of an object's motion and is calculated by multiplying its mass by its velocity. Impulse, on the other hand, is the change in an object's momentum and is calculated by multiplying the force applied to the object by the time it is applied.

2. How is final velocity calculated using momentum and impulse?

The final velocity can be calculated by dividing the impulse applied to an object by its mass. This is because the impulse represents the change in momentum, and dividing by the mass gives us the change in velocity. The formula is vf = J/m, where vf is the final velocity, J is the impulse, and m is the mass of the object.

3. Can momentum and impulse be negative?

Yes, both momentum and impulse can be negative. A negative momentum indicates that the object is moving in the opposite direction of its positive momentum, while a negative impulse indicates that the force applied is in the opposite direction of the object's motion.

4. How does conservation of momentum apply to collisions?

Conservation of momentum states that in a closed system, the total momentum before and after a collision remains the same. This means that the total momentum of all objects involved in the collision will be the same before and after the collision, even if the individual object's momentum changes.

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

Momentum and impulse are important concepts in many fields, including sports, transportation, and engineering. In sports, understanding momentum and impulse can help improve performance and prevent injuries. In transportation, they are crucial for designing safe and efficient vehicles. In engineering, they are used to analyze the impact of forces on structures and machinery.

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