# A ball with a mass of 4 kg moving at 3 m/s hits a still ball with a mass of 26 kg. If the first ball stops moving, how fast is the second ball moving? How much kinetic energy was lost as heat in the collision?

$\frac{6}{13} = 0.461 \setminus \setminus \textrm{\frac{m}{s}}$ & loss of K.E. $= 15.231 \setminus J$

#### Explanation:

Ball of mass ${m}_{1} = 4 k g$ moving with ${u}_{1} = 3 \setminus \frac{m}{s}$ strikes another ball of mass ${m}_{2} = 26 \setminus k g$ at rest ${u}_{2} = 0$. After collision, first ball stops ${v}_{1} = 0$ & second ball moves with the velocity ${v}_{2}$ then by law of conservation of momentum we have

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

$4 \left(3\right) + 26 \left(0\right) = 4 \left(0\right) + 26 {v}_{2}$

${v}_{2} = \frac{12}{26}$

$= \frac{6}{13} \setminus \setminus \textrm{\frac{m}{s}}$

Loss of kinetic energy

$= \setminus \textrm{\in i t i a l K . E .} - \setminus \textrm{f \in a l K . E .}$

$= \frac{1}{2} {m}_{1} {u}_{1}^{2} - \frac{1}{2} {m}_{2} {v}_{2}^{2}$

$= \frac{1}{2} \left(4\right) {\left(3\right)}^{2} - \frac{1}{2} \left(26\right) {\left(\frac{6}{13}\right)}^{2}$

$= 15.231 \setminus J$

Jul 28, 2018

The speed is $= 0.46 m {s}^{-} 1$ and the loss in kinetic energy is $= 15.23 J$

#### Explanation:

There is conservation of momentum

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

The mass ${m}_{1} = 4 k g$

The velocity ${u}_{1} = 3 m {s}^{-} 1$

The mass ${m}_{2} = 26 k g$

The velocity ${u}_{2} = 0 m {s}^{-} 1$

The velocity ${v}_{1} = 0 m {s}^{-} 1$

Therefore,

$4 \cdot 3 + 26 \cdot 0 = 4 \cdot 0 + 26 \cdot {v}_{2}$

${v}_{2} = \frac{12}{26} = 0.46 m {s}^{-} 1$

The loss in kinetic energy is

$\Delta K E = K {E}_{1} - K {E}_{2}$

$= \frac{1}{2} {m}_{1} {u}_{1}^{2} - \frac{1}{2} {m}_{2} {v}_{2}^{2}$

$= \frac{1}{2} \cdot 4 \cdot {3}^{2} - \frac{1}{2} \cdot 26 \cdot {0.46}^{2}$

$= 15.23 J$