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

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Homework Help Overview

The problem involves calculating the change in momentum of a 4.8 kg mass projected at a 55° angle, which lands after 3.6 seconds. The context includes considerations of momentum conservation and the effects of vertical and horizontal components of velocity.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the conservation of momentum along the horizontal axis and question what causes a change in momentum. There are attempts to relate the vertical component of velocity to the change in momentum, with some participants expressing confusion over their calculations.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the conservation of momentum and the behavior of vertical velocity, but there is no explicit consensus on the correct approach or calculations.

Contextual Notes

Participants are working under the assumption of ignoring air resistance, which affects their calculations of momentum change. There are indications of confusion regarding the application of equations and the interpretation of results.

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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