physics help URGENT!!!

unclebutcher SG

An artificial satellite of earth releases a bomb. neglecting air resistance, the bomb will:

hit earth under/ahead/behind the satellite or never hit earth at all

a 0.25 KG block oscillates at the end of the spring with a spring constant of 200N/m . if the oscillation is started by elongating the spring 0.15m and giving the block a speed of 3.0 m/s, the max speed of the block is?

which statement is incorrect for a transverse wave traveling on a string?
aThe energy of the wave is proportional to the square of the waves amplitude

bThe speed of a moving point on the string is the same as the wave speed

cThe speed of the wave increase with frequency and wavelength of the waves

dAll the particles on the string vibrate with the same frequency.

unclebutcher SG

bump!!!!!!!!!!

<Precious> SG

Stooo-pid questions leh.... I want better questions before I answer! Laughing

hisoka SG

i think never hit if i remember correctly. assuming release is just let go.

erm i forgot the equatin. but i think F= kx so can find the force then can find initial acceleration due to F= ma. having boundary conditions that t= 0 v = 3 i think can find max speed.

forgot whats a transverse wave Laughing

eagle SG

uncle u solved already a not? I just realised I missed out seeing this question yesterday

unclebutcher SG

haha actually no....can help? thanks.

eagle SG

Geostationary? Wink

I dunno if your spring is hanging down or horizontal. I shall assume it as horizontal for easier calculations

energy stored by spring at extension of 0.15m = 1/2 * 200 * 0.15^2 = 2.25J
Max speed at 0 extension

using conservation of energy
1/2 m v^2 = 2.25J + 1/2 m 3^2
v = sqrt(27) m/s
not sure about this...

B is wrong because the speed of the moving point varies whereas the wave speed is a constant. The point moves perpendicular to the direction of the wave.

unclebutcher SG

Originally posted by eagle:
I dunno if your spring is hanging down or horizontal. I shall assume it as horizontal for easier calculations

energy stored by spring at extension of 0.15m = 1/2 * 200 * 0.15^2 = 2.25J
Max speed at 0 extension

using conservation of energy
1/2 m v^2 = 2.25J + 1/2 m 3^2
v = sqrt(27) m/s
not sure about this...

how come got extra 1/2?

^tamago^ SG

By conservation of energy,

½mv²=½kx²+½mu²
v²=kx²/m+u²
= [(200N/m)(0.15m)²]/(0.25kg) + (3.0m/s)²
= 27m²/s²
v=3*sqrt(3) m/s

eagle SG

Energy stored in spring is 1/2 k e^2 right?

unclebutcher SG

oh yeah i forgot liao