Calculate Change in Momentum of 4.8 kg Mass Launched at 55°

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SUMMARY

The discussion centers on calculating the change in momentum of a 4.8 kg mass projected at a 55° angle, which lands after 3.6 seconds. The correct approach involves recognizing that while horizontal momentum remains constant, the vertical momentum changes due to gravitational acceleration. The final momentum calculation should consider the vertical component's change, leading to a total change in momentum of 277.44 kg m/s, factoring in the direction of the velocity upon impact.

PREREQUISITES
  • Understanding of basic physics concepts such as momentum and projectile motion.
  • Familiarity with the equations of motion under constant acceleration.
  • Knowledge of vector components, particularly in relation to angles and forces.
  • Ability to apply the impulse-momentum theorem (J = Δp).
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  • Review the principles of projectile motion and its equations.
  • Study the impulse-momentum theorem in detail.
  • Learn how to decompose vectors into their horizontal and vertical components.
  • Explore examples of momentum conservation in different physical scenarios.
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of projectile motion and momentum calculations in a frictionless environment.

omc1
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Homework Statement

An object of mass 4.8 kg is projected into the air at a 55° angle. It hits the ground 3.6 s later. What is the magnitude of its change in momentum while it is in the air? Ignore air resistance

Homework Equations

J=p =sumF Δt=mΔv


The Attempt at a Solution

mvsine(theta) = mgt v=43.07 m/s p=mv = 206.74 kg m/s but its says that's wrong i tried it a dif way but got the same answer please help...
 
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hello omc1 ,
If you notice momentum will remain conserved along horizontal axis . ie component of velocity along horizontal axis will remain same throughout .
So it will not cause any change in momentum .
Then what will cause a change in momentum ?
When you project the mass upward , the vertical component of mass will vary , and you might also know it will come back with the same speed with which it was thrown .
Try to find out this change
 
so msin(theta)gt=p ??
 
twice of that
 
when i did that it is still giving me a wrong answer ...277.44
 
With no air resistance, horizontal speed doesn't change and and vertical speed, once the object has hit the ground, will have the same magnitude but opposite sign.
 
the neg of that answer doesn't work either...
 

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