A ball with a mass of #4 kg # and velocity of #9 m/s# collides with a second ball with a mass of #7 kg# and velocity of #- 5 m/s#. If #80%# of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

This is a relatively complicated collision problem. You won't see many of these unless you're majoring in physics or are an engineer.

By definition, energy is not conserved in this collision, due to the question's statement about kinetic energy being lost.

Therefore, the collision is inelastic and the conservation of momentum is observed, still.

However, given your data we can make a handy approximation,

#1/2m_"A"nu_"A"^2+1/2m_"B"nu_"B"^2 = 1/2m_"A"nu_"A"^('2)+1/2m_"B"nu_"B"^('2)+Q#

Where,

#Q = 0.80"KE"#

Let's start by finding the initial kinetic energy, and recall,

#"KE" = 1/2mnu^2#
#Sigma"KE" approx 250"J"#

So,

#Q = 0.80"KE" approx 200"J"#, and by extension
#"KE"^' = "KE" - Q = 50"J"#

Intuitively, the larger ball will move slower than the faster ball after the collision, even if it is inelastic.

We have two variables, so we need two equations!

#1/2m_"A"nu_"A"^('2)+1/2m_"B"nu_"B"^('2) = 50"J" " "(1)#

Now, remember what we said about conservation of momentum, the momentum after the collision will be the same as before, regardless of the energy lost.

#m_"A"nu_"A" + m_"B"nu_"B" = 1"N"*"m" = m_"A"nu_"A"^' + m_"B"nu_"B"^'#
#therefore m_"A"nu_"A"^' + m_"B"nu_"B"^' = 1"N"*"m" " " (2)#
From here on we will use some substitution by defining #nu_"B"^'# using #(2)#,
#nu_"B" = (1"N" * "m" - 4"kg" * nu_"A")/(7"kg")#
Now, let's solve for #nu_"A"#, and from here on it's smooth sailing,
#1/2 * 4"kg" * nu_"A"^('2) + 1/2 * 7"kg" * ((1"N" * "m" - 4"kg" * nu_"A")/(7"kg"))^2 =50"J"#
#therefore nu_"A"^' approx (4.08"m")/"s"#, and by extension,
#m_"A"nu_"A"^' + m_"B"nu_"B"^' = 1"N"*"m"#
#therefore nu_"B"^' approx (-2.19"m")/"s"#

Thanks to this answer for providing a push in the right direction for me!

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Answer 2

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

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

Using the conservation of momentum:

[ m_1v_1 + m_2v_2 = m_1u_1 + m_2u_2 ]

where ( m_1 ) and ( m_2 ) are the masses of the first and second balls, and ( u_1 ) and ( u_2 ) are their initial velocities.

Substituting the given values:

[ (4 \times 9) + (7 \times (-5)) = (4 \times 9) + (7 \times (-5)) ]

[ 36 - 35 = 36 - 35 ]

[ 1 = 1 ]

This equation is satisfied, indicating that momentum is conserved.

Now, using the 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 ]

Given that 80% of the kinetic energy is lost:

[ \frac{1}{5} (m_1 u_1^2 + m_2 u_2^2) = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 ]

Substituting the given values:

[ \frac{1}{5} \left(4 \times 9^2 + 7 \times (-5)^2\right) = \frac{1}{2} \times 4 \times v_1^2 + \frac{1}{2} \times 7 \times v_2^2 ]

[ \frac{1}{5} (324 + 175) = 2v_1^2 + \frac{7}{2} v_2^2 ]

[ \frac{499}{5} = 2v_1^2 + \frac{7}{2} v_2^2 ]

[ \frac{499}{10} = v_1^2 + \frac{7}{4} v_2^2 ]

Given that 80% of the kinetic energy is lost, 20% is retained. So, we can rewrite the equation as:

[ \frac{1}{5} (m_1 u_1^2 + m_2 u_2^2) = \frac{1}{5} (m_1 v_1^2 + m_2 v_2^2) ]

[ 324 + 175 = 4v_1^2 + 7v_2^2 ]

[ 499 = 4v_1^2 + 7v_2^2 ]

Now, we have two equations:

[ 1. \quad 4v_1 + 7v_2 = 36 - 35 = 1 ]

[ 2. \quad 499 = 4v_1^2 + 7v_2^2 ]

Solving these equations simultaneously will give us the final velocities of the balls.

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Answer from HIX Tutor

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.

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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