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

2.13 m/s

Total mass is 6 kg. Compute the velocity:

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To find the final velocities of the balls after the collision, you can use the principle of conservation of momentum and conservation of kinetic energy.

Let ( v_1 ) and ( v_2 ) be the final velocities of the first and second balls, respectively.

Using conservation of momentum: ( m_1v_1 + m_2v_2 = m_1u_1 + m_2u_2 )

( 4 \times 3 + 2 \times (-1) = 4 \times v_1 + 2 \times v_2 )

( 12 - 2 = 4v_1 + 2v_2 )

( 10 = 4v_1 + 2v_2 ) ...(1)

Using conservation of kinetic energy: ( \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 )

( \frac{1}{2} \times 4 \times 3^2 + \frac{1}{2} \times 2 \times (-1)^2 = \frac{1}{2} \times 4 \times v_1^2 + \frac{1}{2} \times 2 \times v_2^2 )

( 18 + 1 = 2v_1^2 + v_2^2 )

( 19 = 2v_1^2 + v_2^2 ) ...(2)

From equation (1), we can express ( v_2 ) in terms of ( v_1 ):

( v_2 = \frac{10 - 4v_1}{2} )

Substitute this expression for ( v_2 ) into equation (2) and solve for ( v_1 ):

( 19 = 2v_1^2 + \left(\frac{10 - 4v_1}{2}\right)^2 )

Solve this equation to find ( v_1 ). Once you have ( v_1 ), you can find ( v_2 ) using the equation for ( v_2 ) in terms of ( v_1 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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