Doubts on impulse and momentum

In summary: You should be able to solve for them using the equation you have. The equation should be set to 0, so you have:0=(m1+m2)16-20m1Then you could solve for m2 in terms of m1 and plug that into the other equation to solve for m1. I hope that helps.In summary, the first part involves an antitank missile exploding into two pieces and calculating the speed of the first piece based on the given information. The correct answer is v = -56 m/s. In the second part, the mass of each body is being calculated based on the given information and the equation \Delta P = mv_f - mv_o = (m_1 + m_
  • #1
catala
5
0

Homework Statement




  1. An antitank missile (m = 12kg.), when it reaches the maximum height (v = 0) explodes breaking into a large piece of 5kg. and a multitude of small pieces exiting vertically upwards with a speed of 40m / s. What is the speed of the first piece?.


  • A locomotive traveling at 20m / s, collides and engages in a car traveling initially in repose after the two at a speed of 16m / s. if the amount of movement of the locomotive-car system is 128kg.m / s. What is the mass of each body?

Homework Equations



[itex]\Delta P = I [/itex]

[itex]p = m*v[/itex]

The Attempt at a Solution



1. [itex]\Delta P = mv_f -mv_o = (5 + m)*v - (m*0) [/itex]

[itex]= (5+m)v_f = 5*v_f + m*v_f = 5v + 40*7 \to \boxed{v = -56 m/s} [/itex]

2. [itex]\Delta P = mv_f - mv_o = (m_1 + m_2)16 -20m_1 [/itex]

[itex](m_1 + m_2)16 = 128 \to m_1 = \frac{128 -16m_2}{16} [/itex]Could you correct my mistakes?

Thanks
 
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  • #2
I looked at the first part:
v=-56 m/s looks like the right answer to me. The variables in the work get mixed up. v becomes vf then goes back to v. Some of the problem might be the starting equation. There should be a summation for the m*vf term because m breaks up into two parts. So you could have m*vf=(m1*v1f+m2*v2f). Then it would help to state that delta P is 0, and you should still get -56 m/s.
Does that help any?
 
  • #3
For the second question, you have not arrived at values for m1 and m2.
 

1. What is the difference between impulse and momentum?

Impulse and momentum are both important concepts in physics, particularly in relation to the motion of objects. Momentum is a measure of an object's mass and velocity, while impulse is the change in momentum caused by a force acting on an object. In other words, momentum is a property of an object, while impulse is a result of external forces acting on the object.

2. How do you calculate impulse and momentum?

Impulse is calculated by multiplying the force acting on an object by the time for which it acts. This can be represented by the equation J = F * Δt. Momentum, on the other hand, is calculated by multiplying an object's mass by its velocity. This can be expressed as p = m * v.

3. What is the conservation of momentum?

The conservation of momentum is a fundamental principle in physics which states that the total momentum of a system remains constant, unless acted upon by external forces. This means that in a closed system, the total momentum before an event is equal to the total momentum after the event.

4. How does impulse and momentum relate to collisions?

In collisions, impulse and momentum are important factors in determining the outcome of the event. During a collision, the total momentum of the system is conserved, but the impulse from external forces can cause changes in the individual momentum of objects involved. This can result in changes in the objects' velocities or directions of motion.

5. Can impulse and momentum be negative?

Yes, both impulse and momentum can be negative. A negative impulse can occur when a force acts in the opposite direction of an object's motion, resulting in a decrease in momentum. Negative momentum can also occur when an object is moving in the opposite direction of a chosen positive direction, resulting in a negative value for momentum.

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