# A neutron of mass 1.7 x 10^-27 kg, travelling at 2.7 km/s, hits a stationary lithium nucleus of mass 1.2 x 10^-26 kg. After the collision, the...collision, the velocity of the lithium nucleus is...

A neutron of mass 1.7 x 10^-27 kg, travelling at 2.7 km/s, hits a stationary lithium nucleus of mass 1.2 x 10^-26 kg. After the collision, the...

collision, the velocity of the lithium nucleus is 0.40 km/s at 54 degree to the original direction of motion of the neutron. If the speed of the neutron after the collision is 2.5 km/s, in what direction is the neutron now travelling?

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### 1 Answer

A neutron of mass 1.7 x 10^-27 kg, traveling at 2.7 km/s, hits a stationary lithium nucleus of mass 1.2 x 10^-26 kg. After the collision, the collision, the velocity of the lithium nucleus is 0.40 km/s at 54 degree to the original direction of motion of the neutron. The speed of the neutron after the collision is 2.5 km/s.

The law of conservation of momentum states that the total momentum of the system if conserved.

The initial momentum of the system is 1.7*10^-27*2700 kg*m/s

After collision let the neutron move in a direction making an angle x with its original direction. This gives:

0 = 400*1.2 x 10^-26*sin 54 + 1.7*10^-27*2500*sin x

=> sin x = -(400*1.2 x 10^-26*sin 54)/(1.7*10^-27*2500)

sin x = -0.7614

=> x = -49.59 degrees

**The neutron moves in a direction that makes an angle of 49.59 degrees with its original direction but on the opposite side after collision.**