Answer:
Speed of lighter particle = -21.53 m/s
Speed of other particle = 3.8 m/s
Step-by-step explanation:
Let mass of the lighter object be m
Thus mass of heavier object = 4m
Speed of lighter particle = 19 m/s
Speed of second particle with opposite direction = - ⅓(19) m/s = -19/3 m/s
Now, from the formulas of momentum before collision = momentum after collision, and also kinetic energy before collision = Kinetic Energy after collision, we have;
v_bf = [2m_a/(m_a + m_b)]v_ai + [(m_b - m_a)/(m_a + m_b)]v_bi
Now, in this question;
m_a = m
m_b = 4m
v_ai = 19
v_bi = -19/3 m/s
Thus;
v_bf = [2m/(m + 4m)]19 + [(4m - m)/(m + 4m)](-19/3)
Simplifying to get;
v_bf = 19(2m/5m) - (19/3)(3m/5m)
>> v_bf = 38/5 - 19/5
>> v_bf = 19/5 m/s
>> v_bf = 3.8 m/s
Similarly;
v_af = [(m_a - m_b)/(m_a + m_b)]v_ai + [2m_b/(m_a + m_b)]v_bi
v_af = 19((m - 4m)/(m + 4m)) - (19/3)((2 × 4m)/(m + 4m))
This gives;
v_af = 19(-3m/5m) - (19/3)(8m/5m)
v_af = -(57/5) - (152/15)
v_af = -323/15
v_af = -21.53 m/s