A #5# #kg# red toy car is moving right at #10# #ms^-1# when it collides with another #10# #kg# green car moving left at #10# #ms^-1#. What is the velocity of the red car after the collision?

1 Answer
ali ergin
May 31, 2017

#"Animation is the answer to your problem, check it out."#

Explanation:

enter image source here

  • In the solution, the collision is supposed to be completely flexible and central.
  • Remember that velocity and momentum is the quantities of the vector.
  • You should pay attention to the speed signs of moving objects in different directions
  • The red ball seen at the animation symbolizes the red toy car.
  • We solve the momentum problems by using momentum conservation and laws of kinetic energy conservation.
  • I'll give you two methods.

#1)#

#v_r'=(2*Sigma P_b)/((m_r+m_b))-v_r#

#v_b'=(2*Sigma P_b)/((m_r+m_b))-v_b#

#"where ;"#

#m_r:"mass of the red toy car"#
#m_b:"mass of the blue toy car"#
#v_r':"red toy car's velocity after collision"#
#v_r:"red toy car's velocity before collision"#
#v_b':"blue toy car's velocity after collision"#
#v_b:"blue toy car's velocity before collision"#
#P_b:"total momentum before collision"#

#"Let us calculate total momentum before collision."#

#Sigma P_b=m_r*v_r+m_b*v_b#
#Sigma P_b=5*10-10*10#
#Sigma P_b=50-100#
#Sigma P_b=-50 kg*m*s^(-1)#

#v_r'=(2*(-50))/((5+10))-10#

#v_r'=(-100)/(15)-10=(-250)/15#

#v_r'=-16.67" " m*s^(-1)#

#v_b'=(2*(-50))/((15))-(-10)#

#v_b'=((-100))/((15))+10#

#v_b'=(-100+150)/15=50/15#

#v_b'=3.33" "m*s^(-1)#

#2)#

  • Let's write momentum conservation equation.

#m_r*v_r+m_b*v_b=m_r*v_r'+m_b*v_b'#

#5*10-10*10=5*v_r'+10*v_b'#

#50-100=5*v_r'+10*v_b'#

#-cancel(50)=cancel(5)*v_r'+cancel(10)*v_b'#

#-10=v_r'+2v_b'" (1)"#

  • With algebraic operations using the momentum conservation equation and the kinetic energy conservation equation, you can find the following equation.

#v_r+v_r'=v_b+v_b'" (proof can be done.) (2)"#

#10+v_r'=-10+v_b'#

#color(blue)(v_b^'=20+v_r')#

#"we can use the equation (1)"#

#-10=v_r'+2(color(blue)(20+v_r'))#

#-10=v_r'+40+2v_r'#

#-50=3v_r^'#

#v_r^'=-50/3=-16.67 " "m*s^(-1)#

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