# If we were colliding a Ball (182.20 g Mass), at 100% speed (1.94 m/s), into a tennis ball of the same size, but half the mass of the primary ball, what could we determine about the motion and energy of the tennis ball after the collision?

Jun 14, 2018

V= 1.29ms¯¹

$\text{Energy of tennis ball} = 6.9381 \times {10}^{-} 3 J$

#### Explanation:

$\text{Mass of ball} = {M}_{1} = 0.1822 k g$

$\text{Mass of tennis is half of Mass of ball}$

Therefore;

$\text{Mass of tennis} = {M}_{2} = 0.0911 k g$

$\text{Velocity of ball} = {U}_{1} = 1.94 m {s}^{-} 1$

$\text{Velocity of tennis} = {U}_{2} = 0$

Recall, since its inelastic collision..

${M}_{1} {U}_{1} + {M}_{2} {U}_{2} = \left({M}_{1} + {M}_{2}\right) V$

Plugging the values into the equation..

$0.1822 \times 1.94 = \left(0.1822 + 0.0911\right) V$

$V = \frac{0.3535}{0.2733}$

V= 1.29ms¯¹

Then energy of tennis ball..

Using;

$K . E = \frac{1}{2} m {v}^{2}$

Energy of tennis ball $= \frac{1}{2}$ mass of tennis ball $\times {v}^{2}$

$\text{Energy of tennis ball} = \frac{1.672 \times {\left(0.0911\right)}^{2}}{2}$

$\text{Energy of tennis ball} = \frac{1.672 \times 0.00829921}{2}$

$\text{Energy of tennis ball} = \frac{0.01387627912}{2}$

$\text{Energy of tennis ball} = 6.9381 \times {10}^{-} 3 J$