A proton of mass 1.67x10^-27kg, travelling with a speed of 1.0x10^7m/s, collides with a helium nucleus at rest. The proton rebounds straight back with a speed of 6.0x10^6m/s while the helium nucleus moves forward with a speed of 4.0x10^6m/s. What was the momentum of the helium nucleus after the collision?

Question

beth12345 · Answer

$$2.67 \times 10^{-20} kg \times m/s$$ is the answer I'm given, I just don't know how to get it.

anonymous · Answer

mass of helium nucleus=4 amu
                                  =4*1.67*10^-27 kg
                                   =6.68*10^-27 kg
linear momentum=mass *velocity=(6.68*10^-27)*4*10^6 kg-m/sec
                                                 =2.672*10^-20 kg-m/sec

beth12345 · Answer

I don't know why it's multiplied by 4, and im also not sure what the amu stands for

anonymous · Answer

momentum will be conserved, which is the nature of all collisions. 

Therefore, $$p_p = p_p' + p_n'$$ where the prime indicates the momentum after the collision. 

This can be expanded as$$v_p m_p = v_p' m_p + p_n'$$We can solve this for $p_n'$ without even having to know the mass of the nucleus! 

Remember that since the proton rebounds, $v_p' \lt 0$

beth12345 · Answer

ok

beth12345 · Answer

thanks :)