# A ball with a mass of #4# #kg# and velocity of #5# #ms^-1# collides with a second ball with a mass of #6# #kg# and velocity of #-1# #ms^-1#. If #20%# of the kinetic energy is lost, what are the final velocities of the balls?

Momentum is conserved. Kinetic energy is not, but in this case we know that final

The final velocity of the

Prior to the collision:

vigor:

Energy in motion:

Following the collision:

vigor:

Energy in motion:

We are left with two equations involving two unknowns:

Putting this in place of Equation 2:

It's just algebra, so I'll leave you to solve this equation, which involves solving a quadratic equation.

Assume that the "+" direction is from left to right and the "-" direction is from right to left in order to determine which set of answers we want.

Physically speaking, the other set of answers is nonsensical because it assumes that the left-hand ball is moving more quickly to the right than the right-hand ball.

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Before the collision

Momentum:

Kinetic energy:

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To solve for the final velocities of the balls after the collision, we can use the principle of conservation of momentum and the concept of kinetic energy loss.

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

According to the conservation of momentum:

( m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_1 + m_2 \cdot v_2 )

where ( m_1 ) and ( m_2 ) are the masses of the first and second balls, and ( v_{1i} ) and ( v_{2i} ) are their initial velocities.

Given: ( m_1 = 4 ) kg, ( v_{1i} = 5 ) m/s, ( m_2 = 6 ) kg, ( v_{2i} = -1 ) m/s.

Applying the conservation of momentum:

( 4 \cdot 5 + 6 \cdot (-1) = 4 \cdot v_1 + 6 \cdot v_2 )

( 20 - 6 = 4v_1 + 6v_2 )

( 14 = 4v_1 + 6v_2 ) ...(1)

Now, to consider the kinetic energy loss, let ( KE_i ) be the initial kinetic energy and ( KE_f ) be the final kinetic energy.

Given that 20% of kinetic energy is lost, ( KE_f = 0.8 \times KE_i ).

( KE_i = \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 )

( KE_f = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 )

Substituting the given values:

( 0.8 \times (\frac{1}{2} \cdot 4 \cdot 5^2 + \frac{1}{2} \cdot 6 \cdot (-1)^2) = \frac{1}{2} \cdot 4 \cdot v_1^2 + \frac{1}{2} \cdot 6 \cdot v_2^2 )

( 0.8 \times (50 + 3) = 2v_1^2 + 3v_2^2 )

( 0.8 \times 53 = 2v_1^2 + 3v_2^2 )

( 42.4 = 2v_1^2 + 3v_2^2 ) ...(2)

Now, we have two equations (equation 1 and equation 2) with two unknowns ( v_1 ) and ( v_2 ). We can solve them simultaneously to find the final velocities of the balls.

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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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