# A ball with a mass of #17 kg# moving at #9 m/s# hits a still ball with a mass of #21 kg#. If the first ball stops moving, how fast is the second ball moving? How much kinetic energy was lost as heat in the collision?

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To find the velocity of the second ball after the collision, we can use the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision, assuming no external forces act on the system. The momentum ( p ) is given by ( p = mv ), where ( m ) is the mass and ( v ) is the velocity.

Before the collision: ( p_{\text{before}} = m_1 \times v_1 + m_2 \times v_2 )

After the collision: ( p_{\text{after}} = 0 \times v_1 + (m_1 + m_2) \times v_2 )

Using the conservation of momentum principle, we can equate ( p_{\text{before}} ) to ( p_{\text{after}} ) and solve for ( v_2 ).

( m_1 \times v_1 + m_2 \times v_2 = (m_1 + m_2) \times v_2 )

( 17 \times 9 + 21 \times 0 = (17 + 21) \times v_2 )

( 153 = 38 \times v_2 )

( v_2 = \frac{153}{38} )

( v_2 = 4.026 ) m/s (approximately)

To find the kinetic energy lost as heat in the collision, we can use the principle of conservation of kinetic energy. The initial kinetic energy (( KE_{\text{initial}} )) is equal to the final kinetic energy (( KE_{\text{final}} )) plus the heat generated (( Q )).

( KE_{\text{initial}} = KE_{\text{final}} + Q )

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

Given ( m_1 = 17 ) kg, ( v_1 = 9 ) m/s, ( m_2 = 21 ) kg, and ( v_2 = 4.026 ) m/s (as calculated above), we can solve for ( Q ).

( \frac{1}{2} \times 17 \times 9^2 = \frac{1}{2} \times 21 \times (4.026)^2 + Q )

( 688.5 = 214.641 + Q )

( Q = 473.859 ) J (approximately)

Therefore, the second ball is moving at approximately ( 4.026 ) m/s after the collision, and ( 473.859 ) J of kinetic energy was lost as heat in the collision.

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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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