A ball with a mass of $1 k g$ $1 kg$ and velocity of $2 \frac{m}{s}$ $2 m/s$ collides with a second ball with a mass of $3 k g$ $3 kg$ and velocity of $- 4 \frac{m}{s}$ $- 4 m/s$ . If $20 %$ $20%$ of the kinetic energy is lost, what are the final velocities of the balls?

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Answer 1 · 2018-08-13T09:17:13

The solution is $=((-1.03, -3.68))ms^-1$

$m_1u_1+m_2u_2=m_1x+m_2y$

Plugging in the above values

$1*2+3*-4=1x+3y$

$x+3y=2-12=-10$

$x+3y=-10$ ................................. $(1)$

$20%$ of kinetic energy is lost

$(1/2m_1u_1^2+1/2m_2u_2^2)*8/10=1/2m_1x^2+1/2m_2y^2$

Plugging the data

$(1*2^2+3*(-4)^2)*8/10=1x^2+3y^2$

$x^2+3y^2=41.6$

$x^2+3y^2=41.6$ ...................................... $(2)$

Solving equations $(1)$ and $(2)$ graphically

graph{(x+3y+10)(x^2+3y^2-41.6)=0 [-18.02, 18.03, -9.01, 9.01]}

The solutions are $=(-1.03, -3.68)$ or $=(-6.03, -1.32)$

The second solution is discarded

A ball with a mass of 1 kg 1kg1 kg and velocity of 2 m/s2ms2 m/s collides with a second ball with a mass of 3 kg3kg3 kg and velocity of - 4 m/s−4ms- 4 m/s. If 20%20%20% of the kinetic energy is lost, what are the final velocities of the balls?