Question

An unstable nucleus of mass 1.7 ✕ 10−26 kg, initially at rest at the origin of a coordinate system, disintegrates into three particles. One particle, having a mass of m1 = 1.0 ✕ 10−27 kg, moves in the positive y-direction with speed v1 = 5.8 ✕ 106 m/s. Another particle, of mass m2 = 9.0 ✕ 10−27 kg, moves in the positive x-direction with speed v2 = 3.8 ✕ 106 m/s. Find the magnitude and direction of the velocity of the third particle. (Assume that the +x-axis is to the right and the +y-axis is up along the page.) magnitude m/s direction ° counterclockwise from the +x-axis

Answer 1

Answer :

This is a problem dealing with conservation of momentum

since the momentum of the original particle was zero, the sum of momenta of the three particles must be zero

the first particle has y momentum of magnitude:

1.0x10^(-27)x5.8x10^6 m/s = 5.8x10^(-21) kgm/s

the second particle has x momentum of magnitude

9.0x10^(-27)kg x 3.8x10^6m = 3.24x10^(-20) kgm/s

so the third particle must have -y momentum in the same magnitude of the first particle and - x momentum equal to the mag of the second particle

the mass of the final particle is:

1.7x10^(-26)-1.0x10^(-27)-9.0x19^(-27) =
7.0x10^(-27) kg

therefore, its velocity in the -y direciton is:

-5.8x10^(-20)kgm/s / 7.0x10^(-27)kg =- 8.28x10^6m/s

and its velocity in the - x direction is

-3.24x10^(-20)kgm/s/7.0x10^(-27)kg = -4.62 x10^6m/s

the magnitude of vel is

sqrt[(8.28x10^6)^2+(4.62x10^6)^2] =
9.48x10^6m/s

in the direction given by

tan theta = -8.28/-4.62 = 1.79

or 60.80 deg below the negative x axis