# A ball with a mass of 5 kg moving at 2 m/s hits a still ball with a mass of 3 kg. If the first ball stops moving, how fast is the second ball moving?

$3.333 \setminus \setminus \textrm{\frac{m}{\sec}}$

#### Explanation:

One ball of mass ${m}_{1} = 5$kg moving at ${u}_{1} = 2 \setminus \frac{m}{s}$ hits another ball of mass ${m}_{2} = 3$ kg at rest ${u}_{2} = 0$. After collision, the first ball stops i.e. ${v}_{1} = 0$ & second ball moves with a velocity of ${v}_{2}$ in the initial direction of first ball then

By the conservation of momentum in moving direction of first ball

$\setminus \textrm{m o m e n t u m b e f \mathmr{and} e c o l l i s i o n} = \setminus \textrm{m o m e n t u m a f t e r c o l l i s i o n}$

${m}_{1} {u}_{1} + {m}_{2} {u}_{2} = {m}_{1} {v}_{1} + {m}_{2} {v}_{2}$

$5 \left(2\right) + 3 \left(0\right) = 5 \left(0\right) + 3 {v}_{2}$

$3 {v}_{2} = 10$

${v}_{2} = \frac{10}{3}$

$= 3.333 \setminus \setminus \textrm{\frac{m}{\sec}}$

Jul 5, 2018

$\approx 3.33$ meters per second

#### Explanation:

We use the law of conservation of momentum, which states that:

${m}_{1} {u}_{1} + {m}_{2} {u}_{2} = {m}_{1} {v}_{1} + {m}_{2} {v}_{2}$

where:

• ${m}_{1} , {m}_{2}$ are the masses of the two balls

• ${u}_{1} , {u}_{2}$ are the initial velocities of the two balls

• ${v}_{1} , {v}_{2}$ are the final velocities of the two balls

So, we get:

$5 \setminus \text{kg"*2 \ "m/s"+3 \ "kg"*0 \ "m/s"=5 \ "kg"*0 \ "m/s"+3 \ "kg} \cdot {v}_{2}$

${v}_{2} = \left(10 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{kg""m/s")/(3color(red)cancelcolor(black)"kg}}}}\right)$

$\approx 3.33 \setminus \text{m/s}$