Alberto Kozub: The radius isn't given so i'm hoping this project focuses on Linear momentum conservation .20 * 2 = .5 *a million + .2 *v .4 - .5= .2 * v v= -.05m/s the 1st ball strikes lower back at .5 m/s in accordance to this equation yet bodily visualising this factor seems to no longer be conceivable... so the fault seems to be that the .5 Kg ball won't be ready to pass at a million m/s after collision, except the preliminary speed of the .2 kg ball is larger
Tobie Oshea: Use conservation of momentum to find the velocity, then use conservation of energy (Kinetic to Spring) to find x.
Shena Etulain: Let the velocity of the (block+bullet) is v m/s after the collision,=>By the law of momentum conservation:-=>m1u1+m2u2 = (m1+m2) x v=>10 x 10^-3 x 300 + 0 = (10 x 10^-3 + 2.40) x v=>v = 1.24 m/s=>By the law of energy conservation:-=>PE(spring) = KE(block+bullet)=>1/2kx^2 = 1/2(m1+m2)v^2=>17.2 x (x)^2 = 2.41 x (1.24)^2=>(x) = √0.22=>(x) = 0.465959 m=>(x) = 46.6! 0 cm...Show more
Carlee Tangaro: im sorry but if you need help with this...drop out of physics. dont take math or science again in your life. sorry if that hurts to hear, maybe you can paint or play an instrument?
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