A lighter particle moving with a speed of 19 m /s collides with an object of quadruple its mass moving in the opposite direction with a third of its speed. Assume that the collision is a one-dimensional elastic collision. What will be the speed of both particles after the collision?

Speed of lighter particle = -21.53 m/s

Speed of other particle = 3.8 m/s

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

we need to know the time interval during which the force acts. impulse of force = momentum change