Question

An atomic nucleus initially moving at 432m/s emits an alpha particle in the direction of its velocity, and the new nucleus slows to 346m/s. If the alpha particle has a mass of 2u and the original nucleus has a mass of 222u, what speed does the alpha particle have when it is emitted?

I need the correct answer to give full rating.

Answer 1

Using the conservation of momentum principle-->>The momentum after the emission of alpha must remain the same after the emission.

Since the formula is p=mv, if the mass changes, the speen must change accordingly.

Before, p=mv= 432 x 222 =95,904 kg.m/s

After, you have two particles. The sum of their two momentum must still be 95904 kg.m/s

The remaining nucleus has a mass of 220 (i.e 222u - 2u) and a momentum of:
p=mv= 220 x 346 = 76120 kg.m/s

The alpha particle has a momentum of
95904 - 76120 = 19784 kg.m/s

So you can calculate its speed with
v=p/m = 19784/2 = 9892 m/s. --->>ans

Answer 2

This is a question about the conservation of momentum (mass*velocity). Momentum before event = momentum after event.

Before: 432*222= 95,904

After: 222*346 + 2*v = 76,812 + 2v (v = speed of alpha particle)

2v = 19,960

v = 9,530m/s