A ball with a mass of #4 kg # and velocity of #1 m/s# collides with a second ball with a mass of #2 kg# and velocity of #- 5 m/s#. If #75%# 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 #2 kg# and velocity of #-  5 m/s#. If #75%# of the kinetic energy is lost, what are the final velocities of the balls?

Benjamin Asiedu · Accepted Answer

The final velocities are first ball #=-0.17ms^-1 # and the second ball # v_2=-2.66ms^-1#

By the Law of conservation of momentum

#m_1u_1+m_2u_2=m_1v_1+m_2v_2#

The initial kinetic energy is

#KE_1=1/2m_1u_1^2+1/2m_2u_2^2#

The final kinetic energy is

#KE_2=1/2m_1v_1^2+1/2m_2v_2^2#

And

#KE_2=0.25KE_1#

so,

#4*1-2*5=4v_1+2v_2#

#4v_1+2v_2=-6#

#v_2=-(2v_1+3)#...................................#(1)#

#KE_1=1/2m_1u_1^2+1/2m_2u_2^2#

#=1/2*4*1^2+1/2*2*5^2=2+25=27J#

#KE_2=0.25*27=6.75J#

#1/2m_1v_1^2+1/2m_2v_2^2=6.75#

#4v_1^2+2v_2^2=13.5#

#2v_1^2+v_2^2=6.75#..................................#(2)#

Solving for #v_1# and #v_2# in equations #(1)# and #(2)#

#2v_1+(2v_1+3)^2=6.75#

#2v_1+4v_1^2+12v_1+9-6.75=0#

#4v_1^2+14v_1+2.25=0#

Solving this quadratic equation

#v_1=(-14+-sqrt(14^2-4*4*2.25))/(2*4)#

#=(-14+-sqrt160)/(8)=(-14+-12.65)/8#

#v_1=-3.33ms^-1# or #v_1=-0.17#

And

#v_2=-(2v_1+3)=3.66# or #v_2=-2.66#

The final speeds are #(v_1=-3.33 and v_2=3.66)# or #v_1=-0.17 and v_2=-2.66#

Isabella Alvarez · Answer

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

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

Using conservation of momentum:
$$ m_1 \cdot v_1 + m_2 \cdot v_2 = m_1 \cdot u_1 + m_2 \cdot u_2 $$

And using the formula for 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 75% of kinetic energy is lost:
$$ \frac{1}{2} m_1 \cdot u_1^2 + \frac{1}{2} m_2 \cdot u_2^2 = 0.25 \times (\frac{1}{2} m_1 \cdot v_1^2 + \frac{1}{2} m_2 \cdot v_2^2) $$

You can solve these equations to find the final velocities $ v_1 $ and $ v_2 $.