Physics CIE help

[astarmathsandphysics]

It is 11a) SSince the collision is elastic, separation speed =approach speed
v_b-v_a =u+b+u_a

[3ishakay]

How do we do question number 2)c)

in paper 2 .. may/june 09??

2nd variant ppr

[halosh92]

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

[astarmathsandphysics]

sorry only just noticed this. In morning

[astarmathsandphysics]

2c jun 09 p2
impulse =m(v-u)=1.2(-0.8-4))=-5.76
equal and opposite impulse is exerted on 3.6kg mass
so 5.76=m(v-u)=3.6v so v=3.76/3.6=1.6m/s

[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, u_b is in the negative direction so shouldn't it be u_a-u_b=v_a+v_b?

[astarmathsandphysics]

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

[halosh92]

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

thankyou
but could u please solve the questions i posted on the first page its very important sir 

[astarmathsandphysics]

When I get home

[joel]

Just Remember the following

ALL Collisions : Momentum of the system is conserved

Momentum = mass x Velocity

Elastic Collisions

Kinetic Energy of the system is conserved

Kinetic Energy = 1/2 x m x v^2

[halosh92]

2006 nov
Q6 ...i dont get the whole thing

[vanibharutham]

Paper 2?

[astarmathsandphysics]

q15 C by symmetry horizonatally and cos the long one must be the sum of the other horizontal componesnt

[astarmathsandphysics]

2007 oct/nov
q5a
could someone plz sketch it :SSSS
1/2 the intensity is sqrt(1/2)=0.707 times the amplitude
The sum of the two smaller waves is the bigger one