A ball with a mass of #4 kg # and velocity of #1 m/s# collides with a second ball with a mass of #6 kg# and velocity of #- 5 m/s#. If #10%# of the kinetic energy is lost, what are the final velocities of the balls?

Question

A ball with a mass of #4 kg # and velocity of #1  m/s# collides with a second ball with a mass of #6 kg# and velocity of #-  5 m/s#. If #10%# of the kinetic energy is lost, what are the final velocities of the balls?

Sebastian Acosta · Accepted Answer

4kg particle - #-5.86" m/s"#
6kg particle - #-0.424" m/s"#

Diagrams are great for this stuff!

Let the #4kg# ball have a final velocity of #u#, and the #6kg# have a final velocity of #v#

I'm gonna change the question slightly, in that I'll take the second ball to have #5"m/s"# and the first to have #-1"m/s"#. The maths is exactly the same, but since the balls will probably rebound in the same direction as the 6kg one is travelling, this stops having loads of negatives everywhere.

#"KE before"=1/2xx4xx(-1)^2+1/2xx6xx5^2#
#=77J#

#"KE after"=77xx0.9#
#=69.3J#

#:.69.3=1/2xx4xxu^2+1/2xx6xxv^2#
#69.3=2u^2+3v^2" " (1)#

We can form another equation using the conservation of momentum, and solve simultaneously.

#"Using conservation of momentum"#
#"momentum before"="momentum after"#

#4xx(-1)+6xx5=4u+6v#
#26=4u+6v#
#13=2u+3v#
#2u=13-3v#
#u=13/2-3/2v#

#"sub in eqn" (1)#

#69.3=2(13/2-3/2v)^2+3v^2#
#69.3=2(169/4-39/2v+9/4v^2)+3v^2#
#69.3=169/2-39v+9/2v^2+3v^2#
#15/2v^2-39v+76/5=0#
#75v^2-390v+152=0#

From the quadratic formula:

#v=(390+-sqrt(390^2-4xx75xx152))/(2xx75#
#v=4.78# or #v=0.424#

#u=13/2-3/2v#

Let #v=4.78#
#u=13/2-3/2(4.78)#
#u=-0.67#

However, since the #6kg# particle is in the same direction as its original motion (v is positive and we decided its initial velocity was +5m/s), the direction of motion of the #4kg# particle should also be reversed. So we reject this pair of values.

Let #v=0.424#
#u=13/2-3/2(0.424)#
#=5.864#

So the speed of the 4kg particle is 5.86 m/s, and the speed of the 6kg particle was 0.424 m/s. Since they are travelling, in the context given by the question (we reversed this for our maths, remember?), their final speeds are #-5.86" m/s"# and #-0.424" m/s"# respectively.

Isabella Alvarez · Answer

undefined To calculate the final velocities of the balls after the collision, we can use the principle of conservation of momentum and conservation of kinetic energy. Let's denote the initial velocities of the balls as $u_1$ and $u_2$, and the final velocities as $v_1$ and $v_2$, respectively. According to the conservation of momentum:

$$ m_1 \cdot u_1 + m_2 \cdot u_2 = m_1 \cdot v_1 + m_2 \cdot v_2 $$

where $m_1$ and $m_2$ are the masses of the first and second balls, respectively.

And according to the conservation of kinetic energy:

$$ \frac{1}{2} m_1 \cdot u_1^2 + \frac{1}{2} m_2 \cdot u_2^2 = \frac{1}{2} m_1 \cdot v_1^2 + \frac{1}{2} m_2 \cdot v_2^2 $$

Given that 10% of the kinetic energy is lost, the final kinetic energy is 90% of the initial kinetic energy. This can be expressed as:

$$ \frac{1}{2} m_1 \cdot u_1^2 + \frac{1}{2} m_2 \cdot u_2^2 = 0.9 \left( \frac{1}{2} m_1 \cdot v_1^2 + \frac{1}{2} m_2 \cdot v_2^2 \right) $$

Now, we can substitute the given values and solve for $v_1$ and $v_2$. Plugging in the numbers:

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

$$ \frac{1}{2} \cdot 4 \cdot 1^2 + \frac{1}{2} \cdot 6 \cdot (-5)^2 = 0.9 \left( \frac{1}{2} \cdot 4 \cdot v_1^2 + \frac{1}{2} \cdot 6 \cdot v_2^2 \right) $$

Solving these equations simultaneously will give us the final velocities $v_1$ and $v_2$.