A ball with a mass of #6 kg # and velocity of #5 m/s# collides with a second ball with a mass of #1 kg# and velocity of #- 7 m/s#. If #50%# of the kinetic energy is lost, what are the final velocities of the balls?

Question

A ball with a mass of #6 kg # and velocity of #5  m/s# collides with a second ball with a mass of #1 kg# and velocity of #-  7 m/s#. If #50%# of the kinetic energy is lost, what are the final velocities of the balls?

Nathan Ashcroft · Accepted Answer

The velocities of the balls are #=4.04ms^-1# and #=1.24ms^-1#

Energy is preserved

#m_1u_1+m_2u_2=m_1*v_1+m_2v_2#

#u_1=5ms^-1#

#u_2=-7ms^-1#

#m_1=6kg#

#m_2=1kg#

#6*5-1*7=6v_1+1*v_2#

#6v_1+v_2=23#--------------#(1)#

#KE# before is #=1/2m_1u_1^2+1/2m_2u_2^2#

#=1/2*6*5^2+1/2*1*(-7)^2=75+49/2=99.5#

#KE# after is #=1/2m_1v_1^2+1/2m_2v_2^2#

#=1/2*6v_1^2+1/2*1v_2^2=99.5/2#

Thus,

#6v_1^2+v_2^2=99.5#------------------#(2)#

Two equations with two unknowns each

From #(1)#

#v_2=23-6v_1#

Replacing in #(2)#

#6v_1^2+(23-6v_1)^2=99.5#

#6v_1+529-276v_1+36v^2=99.5#

#42v_1^2-276v_1+429.5=0#

#v_1=(276+-sqrt(276^2-4*42*429.5))/84#

#v_1=(276+-sqrt4020)/84#

#v_1=2.6# or #v_1=4.04#

#v_2=7.4# or #v_2=-1.24#

The velocities of the balls are #=4.04ms^-1# and #=1.24ms^-1#

Isabella Alvarez · Answer

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

1. Conservation of momentum: $m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_{1f} + m_2 \cdot v_{2f}$
2. Equation for kinetic energy: $KE = \frac{1}{2}m_1v_{1}^2 + \frac{1}{2}m_2v_{2}^2$

Using these equations and given data, solve for the final velocities.

After solving, you'll find the final velocity of the first ball to be approximately 1.2 m/s and the final velocity of the second ball to be approximately -0.6 m/s.