Explosive collision initial velocity

  • Thread starter fernancb
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  • #1
fernancb
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Homework Statement


A 3.55 kg object sliding on a frictionless surface breaks into two masses that are both equal to half of the original mass. The velocities of the masses are 3.26 m/s due north and 4.58 m/s, 25.4° north of east. Determine the original speed of the 3.55 kg mass.


Homework Equations


p = mv
Ek = 1/2 mv^2 But this isn't an elastic collision, so the conservation of energy wouldn't work


The Attempt at a Solution


Momentum is conserved.
(I'm assuming the mass is moving in a horizontal direction)

m1x v1x = m2x v2x cos(theta)
(3.55)(v1x) = (3.55/2)(4.58)(cos 25.4)
2(v1x) = (4.58)(cos25.4)
v1x = 2.07 m/s

Erm... I resolved it in terms of the x direction so I'm not sure why it isn't working...
 

Answers and Replies

  • #2
rl.bhat
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Since the fragments are moving in the north and north of east, the object is making some angle with east-west direction.
 
  • #3
fernancb
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But didn't I do that? I took into account the fragment was moving in a North east direction... and then resolved it to get my x-component which corresponded with my initial velocity
 
  • #4
gneill
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If they're asking for the original speed (and not the velocity), do you need to worry about components after you determine the magnitude of the total momentum?
 
  • #5
fernancb
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Goin out on a limb here... no? Because speed isn't a vector. But where did I go wrong?
 
  • #6
gneill
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What are the individual momenta of the two fragments (in terms component vectors)?
 
  • #7
fernancb
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object moving North p=(m/2)v2
object moving north east p=(m/2)v3
 
  • #8
gneill
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object moving North p=(m/2)v2
object moving north east p=(m/2)v3

Put some numbers to them, and express them as x and y components. Due east is the positive x-axis. Due north is the positive y-axis.
 
  • #9
fernancb
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X-axis: (3.55/2)(4.58)(cos25.4)=7.34m/s
Y-axis: (3.55/2)(4.58)(sin25.4)+(3.55/2)(3.26) = 9.27 m/s

Resolve? So I wouldn't need to bother with angles after?

Initial speed: (sqrt(7.34^2)+(9.27^2)) = 11.8 m/s
 
  • #10
fernancb
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I tried 11.8 m/s and it's wrong
 
  • #11
gneill
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Careful, those are momenta that you're calculating, not velocities --- kg*m/s.

Since momentum is conserved, and your total momentum has both x and y components, your earlier assumption about the "horizontal" motion of the original object wasn't true.

But what you can say now is that the original speed must be the magnitude of the momentum over the total mass: u = p/m.
 
  • #12
fernancb
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Oh no! so my initial velocity would be = 11.8/3.55 = 3.32 m/s
 
  • #13
gneill
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Oh no! so my initial velocity would be = 11.8/3.55 = 3.32 m/s

Yup. And if you'd kept more significant figures through the process you'd have 3.33 m/s at the end.:smile:
 

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