A neutron collides elastically with a helium nucleus (at rest initially) whose m

Question

A neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus is observed to rebound at an angle ?'2 = 40° from the neutron's initial direction. The neutron's initial speed is 7.0e5m/s. Determine the angle at which the neutron rebounds, ?'1, measured from its initial direction.
_____°
What is the speed of the neutron after the collision?
______ m/s
What is the speed of the helium nucleus after the collision?
_________ m/s

Explanation / Answer

   Mass of the Nuetron = m1 = 1.67 x10-27kg     Mass of the Helium = 4 (1.67x10-27)kg = 6.68x10-27kg     Intial speed of the Neutron = V1i = 7 x105 m/s     Apply , consrevation of linear momentum     Along X - axis      m1 V1i + 0 = m1 Vif cos + m2 V2f cos40         ( As m2 = 4 m1 )      V1i = Vif cos + 4 V2f cos40       Vif cos = V1i - (3.06 )V2f     ---(1)       Along X - axis       0 = m1 V1f sin - m2 V2f sin 40    m1 V1f sin = - 4 m1 V2f sin 40         V1f sin = - 2.57 V2f    ---(2)            (1) 2 + (2)2           V1f2 = (V1i - (3.06 )V2f ) 2 + (- 2.57 V2f  )2   --(2)         Apply . conservation of energy       1/2 m1 V1i 2 + 0 = 1/2 m1 V1f 2+ 1/2 m2 V2f2                V1i 2 = V1f 2 + V2f2      --(3)               sub . eqn (2) in (3)    V1i 2 = (V1i - (3.06 )V2f ) 2 + (- 2.57 V2f  )2 + V2f2          V1i 2 =  V1i 2 + 9.36 V2f 2   - 2 (V1i )(V2f)    + 6.6V2f 2 +  V2f2       2 (V1i )(V2f) = 16.96 V2f 2                 (V1i )= 8.48(V2f) Thus, V2f = V1i / 8.48                      = (7 x105) / 8.48 = 0.82 x105m/s           From eqn (3)            V1f    = ( V1i 2 - V2f2 )                        = 6.95 x105m/s            from eqn (2)    V1f sin = - 2.57 V2f              sin =-17.65o                 V1f2 = (V1i - (3.06 )V2f ) 2 + (- 2.57 V2f  )2   --(2)         Apply . conservation of energy       1/2 m1 V1i 2 + 0 = 1/2 m1 V1f 2+ 1/2 m2 V2f2                V1i 2 = V1f 2 + V2f2      --(3)               sub . eqn (2) in (3)    V1i 2 = (V1i - (3.06 )V2f ) 2 + (- 2.57 V2f  )2 + V2f2          V1i 2 =  V1i 2 + 9.36 V2f 2   - 2 (V1i )(V2f)    + 6.6V2f 2 +  V2f2       2 (V1i )(V2f) = 16.96 V2f 2                 (V1i )= 8.48(V2f) Thus, V2f = V1i / 8.48                      = (7 x105) / 8.48 = 0.82 x105m/s           From eqn (3)            V1f    = ( V1i 2 - V2f2 )                        = 6.95 x105m/s            from eqn (2)    V1f sin = - 2.57 V2f              sin =-17.65o                           

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