# Explosive collision initial velocity

fernancb

## 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

Homework Helper
Since the fragments are moving in the north and north of east, the object is making some angle with east-west direction.

fernancb
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

Mentor
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?

fernancb
Goin out on a limb here... no? Because speed isn't a vector. But where did I go wrong?

Mentor
What are the individual momenta of the two fragments (in terms component vectors)?

fernancb
object moving North p=(m/2)v2
object moving north east p=(m/2)v3

Mentor
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.

fernancb
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

fernancb
I tried 11.8 m/s and it's wrong

Mentor
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.

fernancb
Oh no! so my initial velocity would be = 11.8/3.55 = 3.32 m/s

Mentor
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. 