A stationary object with mass mB is struck head-on by an object with mass mA tha
ID: 1699103 • Letter: A
Question
A stationary object with mass mB is struck head-on by an object with mass mA that is moving initially at speed v_0.a)& b).If the collision is elastic, what percentage of the original energy does each object have after the collision?
c)& d).What does your answer in parts A and B give for the special case mA = mB?
e)& f).What does your answer in parts A and B give for the special case mA = 5mB?
g).For what values, if any, of the mass ratio mA/mB is the original kinetic energy shared equally by the two objects after the collision?
Explanation / Answer
Given
mass 1 ia mA
mass 2 is mB
initial velocity of mass 1 ia vo
initial velocity of mass 2 is 0 m/s
here the collision is elastic so the momentum and energy is conserved
mA vo = mA vA + mB vB
mA ( vo - vA ) = mB vB ..........(1)
1/2 mA vo^2 = 1 / 2 mA vA ^2 + 1 / mB vB ^2
mA ( vo^2 - vA^2 ) = mB vB ^2 .........(2)
dividing equation (2) by (1)
vo + vA = vB
vA = vB - vo
substitute vA valuie in equation (1)
vB = 2mA vo / ( mA + mB)
vA = (mA - mB ) vo / ( mA + mB)
initial energy of mass 1 is 1/2 mA vo^2
final energy of mass 1 is 1 / 2 mA ( (mA - mB ) vo / ( mA + mB))^2
percentage of the original energy does each object have after the collision
For mass 1 = [ 1/2 mA vo^2 ] / [ 1 / 2 mA ( (mA - mB ) vo / ( mA + mB))^2]
= ( mA + mB / mA - mB )^2 * original energy
for mass 2 percentage of the original energy
= mA + mB / 2mA * original energy
if mA = mB
c) infinity for mass 1
d) same energy for mass2
if mA = 5 mB
e) 2.25 * original energy = 225 %
d) 0.36 * original energy = 36%
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