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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: bully3000 on April 05, 2013, 01:58:03 pm

Title: Energy conversions
Post by: bully3000 on April 05, 2013, 01:58:03 pm

A 65kg boy and his 40kg sister, both wearing roller blades, face each other at rest. The girl pushes the boy hard, sending him backwards with a velocity of 2.9m/s towards the west. Ignore friction. What is the minimum amount of chemical energy that is converted into mechanical energy in the girl's muscles?

Title: Re: Energy conversions
Post by: mark_alec on April 05, 2013, 02:26:50 pm

Things to note in order to solve this:
1) momentum is conserved (hint: use this to calculate the girl's velocity)
2) energy is conserved (hint: use this to calculate the total kinetic energy of the motion)

Title: Re: Energy conversions
Post by: bully3000 on April 05, 2013, 04:30:17 pm

Quote from: mark_alec on April 05, 2013, 02:26:50 pm

Things to note in order to solve this:
1) momentum is conserved (hint: use this to calculate the girl's velocity)
2) energy is conserved (hint: use this to calculate the total kinetic energy of the motion)

Momentum=mass*velocity
Momentum (of boy)=65kg*2.9m/s=188.5kg*m/s

As momentum is conserved:

Momentum (of girl)=188.5kg*m/s
Hence, her velocity can be figured out as well.

188.5kg*m/s=40kg*velocity
Velocity (of girl)=4.7125 m/s

KE=1/2mv^2
=1/2(40kg)(4.7125)^2
=444.153125
=444.15 J

Is this right?

Title: Re: Energy conversions
Post by: bully3000 on April 05, 2013, 04:50:20 pm

Or is it:

m₁v₁+m₂v₂=0
6.5(2.9) + 40(v₂)=0
v₂=-4.7m/s

KE=1/2mv^2
=1/2(40)(-4.7)^2
=441.8 J

Please someone clarify?
Still somewhat slightly confused... :(