Answer
(a) The mass of the nucleus is 9m.
(b) The speed of the nucleus after the collision is $3.0\times 10^6~m/s$.
Work Step by Step
Since the collision is elastic, we can use equation (8.27) to find the final velocity $v_B'$ of the nucleus.
$v_A - v_B = v_B' - v_A'$
$v_B' = v_A - v_B + v_A'$
$v_B' = (1.50\times 10^7~m/s) - 0 + (-1.20\times 10^7~m/s)$
$v_B' = 3.0\times 10^6~m/s$
We can use conservation of momentum to find the mass $M$ of the nucleus.
$p_2 = p_1$
$M~(3.0\times 10^6~m/s) +m~(-1.20\times 10^7~m/s) = m~(1.50\times 10^7~m/s)$
$M~(3.0\times 10^6~m/s) = m~(1.50\times 10^7~m/s)-m~(-1.20\times 10^7~m/s)$
$M = \frac{m~(2.70\times 10^7~m/s)}{(3.0\times 10^6~m/s)}$
$M = 9~m$
(a) The mass of the nucleus is 9m.
(b) The speed of the nucleus after the collision is $3.0\times 10^6~m/s$.
