A ball with a mass of #5 kg # and velocity of #2 m/s# collides with a second ball with a mass of #8 kg# and velocity of #- 3 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 #5 kg # and velocity of #2  m/s# collides with a second ball with a mass of #8 kg# and velocity of #-  3 m/s#. If #75%# of the kinetic energy is lost, what are the final velocities of the balls?

Leo Abeyta · Accepted Answer

#v_1=-0.089\ m/s# & #v_2=-1.694\m/s# OR
#v_1=-2.064\ m/s# & #v_2=-0.46\m/s# Let #u_1=2 m/s# & #u_2=-3 m/s# be the initial velocities of two balls having masses #m_1=5\ kg\ # & #\m_2=8\ kg# moving in opposite directions i.e. first one is moving in +ve x-direction & other in -ve x-direction, After collision let #v_1# & #v_2# be the velocities of balls in +ve x-direction By law of conservation of momentum in +ve x-direction, we have #m_1u_1+m_2u_2=m_1v_1+m_2v_2# #5(2)+8(-3)=5v_1+8v_2# #5v_1+8v_2=-14\ .......(1)# Now, loss of kinetic energy is #75%# hence #(1-\frac{75}{100})(\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2)=(\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2)# #1/4(\frac{1}{2}5(2)^2+\frac{1}{2}8(3)^2)=\frac{1}{2}5v_1^2+\frac{1}{2}8v_2^2# #5v_1^2+8v_2^2=23 \ ......(2)# substituting the value of #v_2=\frac{-5v_1-14}{8}# from (1) into (2) as follows #5v_1^2+8(\frac{-5v_1-14}{8})^2=23# #65v_1^2+140v_1+12=0# solving above quadratic equation, we get #v_1=-0.089, -2.064# & corresponding values of #v_2=-1.694, -0.46# Hence, the final velocities of both the balls are either #v_1=-0.089\ m/s# & #v_2=-1.694\m/s# or #v_1=-2.064\ m/s# & #v_2=-0.46\m/s#

Charles Aeschliman · Answer

undefined The final velocities can be calculated using the law of conservation of linear momentum and the given information. The equation for the conservation of linear momentum is $ m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_{1f} + m_2 \cdot v_{2f} $, where $ m_1 $ and $ m_2 $ are the masses of the first and second balls, $ v_{1i} $ and $ v_{2i} $ are their initial velocities, and $ v_{1f} $ and $ v_{2f} $ are their final velocities.

Additionally, the equation for kinetic energy loss is $ \text{KE loss} = \frac{1}{2} \cdot (m_1 \cdot (v_{1f}^2 - v_{1i}^2) + m_2 \cdot (v_{2f}^2 - v_{2i}^2)) $. Given that 75% of kinetic energy is lost, $ \text{KE loss} = 0.75 \cdot \text{initial KE} $, where $ \text{initial KE} $ is the initial kinetic energy of the system.

By solving these equations simultaneously, the final velocities $ v_{1f} $ and $ v_{2f} $ can be determined.