Question

12. A 0.5 kg ball traveling at 6.2 m/s collides head-on with a 1.0 kg ball traveling in the

Answer 1

The final speed of the 1.0 kg ball is 1.9 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system must be conserved before and after the collision, therefore we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where we have:

$m_1 = 0.5 kg$ is the mass of the first ball

$u_1 = 6.2 m/s$ is the initial velocity of the first ball (taking its direction as positive direction)

$v_1 = -14.0 m/s$ is the final velocity of the first ball (in the opposite direction)

$m_2 = 1.0 kg$ is the mass of the second ball

$u_2 = -12.0$ is the initial velocity of the second ball

$v_2$ is the final velocity of the second ball

By re-arranging the equation and by substituting the values, we find: the final velocity of the second ball:

$v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(0.5)(6.2)+(1.0)(-12.0)-(0.5)(-14.0)}{1.0}=-1.9 m/s$

And since the sign is negative, it means that the second ball travels in the same direction it was initially travelling.

Its speed is therefore (magnitude of the velocity) 1.9 m/s.

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