A ball with a mass of $9 kg$ and velocity of $2 m/s$ collides with a second ball with a mass of $3 kg$ and velocity of $- 5 m/s$ . If $10%$ of the kinetic energy is lost, what are the final velocities of the balls?

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Answer 1 · 2018-07-29T18:59:58

The solution is $=((-1.41, 5.23))ms^-1$

$m_1u_1+m_2u_2=m_1x+m_2y$

Plugging in the above values

$9*2+3*-5=9x+3y$

$9x+3y=18-15=3$

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

$10%$ of kinetic energy is lost

$(1/2m_1u_1^2+1/2m_2u_2^2)*9/10=1/2m_1x^2+1/2m_2y^2$

Plugging the data

$(9*2^2+3*5^2)*9/10=9x^2+3y^2$

$9x^2+3y^2=99.9$

$3x^2+y^2=33.3$ ...................................... $(2)$

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

graph{(3x+y-1)(3x^2+y^2-33.3)=0 [-18.02, 18.03, -9.01, 9.01]}

The solutions are $=(-1.41, 5.23)$ or $=(1.91, -4.73)$

The second solution is discarded

A ball with a mass of 9 kg 9 kg and velocity of 2 m/s2 m/s collides with a second ball with a mass of 3 kg3 kg and velocity of - 5 m/s- 5 m/s. If 10%10% of the kinetic energy is lost, what are the final velocities of the balls?