What Are the Correct Speeds and Masses in These Momentum Problems?

Click For Summary
SUMMARY

The discussion focuses on solving momentum problems involving an antitank missile and a locomotive-car collision. The antitank missile, with a mass of 12 kg, explodes into a 5 kg piece and smaller fragments, with the latter moving upwards at 40 m/s, resulting in a calculated speed of -56 m/s for the first piece. In the locomotive scenario, the momentum of the system is given as 128 kg·m/s, and the equations derived for the masses of the locomotive and car require further clarification and correction to accurately solve for m1 and m2.

PREREQUISITES
  • Understanding of momentum conservation principles
  • Familiarity with the equations of motion, specifically p = mv
  • Basic algebraic manipulation skills
  • Knowledge of explosive dynamics in physics
NEXT STEPS
  • Study the conservation of momentum in explosive events
  • Learn how to apply the impulse-momentum theorem in various scenarios
  • Explore detailed examples of collision problems in physics
  • Review the principles of elastic and inelastic collisions
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and momentum, as well as educators looking for examples of momentum problems in real-world applications.

catala
Messages
4
Reaction score
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
 
Physics news on Phys.org
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?
 
For the second question, you have not arrived at values for m1 and m2.
 

Similar threads

Replies
19
Views
2K
Replies
3
Views
2K
Replies
3
Views
2K
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
7
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
16K