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Physics CIE help

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astarmathsandphysics:
paper 2
cie AS physics 2006 nov
Q 3c ii
thx

conservation of momentum
before decay particle is stationary so momentum=0
hence 0=m_1 v_1 +m_2 v_2
0=4*1.8*10^7 +(208-4)*v
v=-7.2*10^7 /204 =-3.53*10^5m/s

tmisterr:
thanks astar! help me understand though, ub is in the negative direction so shouldn't it be ua-ub=va+vb?

astarmathsandphysics:
If they move together you add the speeds
If the are moving apart you subtract.
Iemphasise APPROACH SPEED AND SEPARATION SPEED

halosh92:

--- Quote from: astarmathsandphysics on April 11, 2010, 08:40:58 am ---paper 2
cie AS physics 2006 nov
Q 3c ii
thx

conservation of momentum
before decay particle is stationary so momentum=0
hence 0=m_1 v_1 +m_2 v_2
0=4*1.8*10^7 +(208-4)*v
v=-7.2*10^7 /204 =-3.53*10^5m/s

--- End quote ---
thankyou
but could u please solve the questions i posted on the first page its very important sir  ;D

astarmathsandphysics:
When I get home

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