(a) Find the velocity of the third particle. (b) Find the total kinetic energy i

Question

(a) Find the velocity of the third particle.

(b) Find the total kinetic energy increase in the process.

An unstable atomic nucleus of mass 1.64 X 10^6 m/s. (a) Find the velocity of the third particle. (b) Find the total kinetic energy increase in the process. X 10^-27 kg, moves in the x direction with a speed of 4.00 X 10^6 m/s. Another particle, of mass 8.42 X 10^-27 kg, moves in the ydirection with a speed of 6.00 X 10-26 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.12

Explanation / Answer

as initital momentum was zero

final momnetum in x and y direction must be zero

final momentum in y direction = 5.12 x 10^-27 x 6 x 10^6 kgm/s

final momentum in x direction = 8.42 x 10^-27 x 4 x 10^6 kgm/s

thus the 3rd particle has velocity with both x and y compoments

mass of the 3rd particle = 1.64 x 10^26 - 13.54 x 10^-27 = 2.86 x 10^-27 kg

x componnt of its velocity = 8.42 x 10^-27 x 4 x 10^6 /  2.86 x 10^-27 = 11.77 x 10^6 towards -x

y component of its velocity = 5.12 x 10^-27 x 6 x 10^6 / 2.86 x 10^-27 = 10.74 x 10^6 m/s

net velocity = root (Vx^2 + Vy^2)

direction = tan ^-1 ( Vy/Vx)

KE = .5 [ m1v1^2 + m2v2^2 + m3v3^2]

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