A ball with a mass of #2 kg# is rolling at #9 m/s# and elastically collides with a resting ball with a mass of #1 kg#. What are the post-collision velocities of the balls?

Question

A ball with a mass of #2  kg# is rolling at #9     m/s# and elastically collides with a resting ball with a mass of #1 kg#. What are the post-collision velocities of the balls?

Leo Abeyta · Accepted Answer

No #cancel(v_1=3 m/s)#
No #cancel(v_2=12 m/s)#

the speed after collision of the two objects are see below fro explanation:
#color(red)(v'_1 = 2.64 m/s, v'_2 = 12.72 m/s)# #"use the conversation of momentum"# #2*9+0=2*v_1+1*v_2# #18=2*v_1+v_2# #9+v_1=0+v_2# #v_2=9+v_1# #18=2*v_1+9+v_1# #18-9=3*v_1# #9=3*v_1# #v_1=3 m/s# #v_2=9+3# #v_2=12 m/s# Because there are two unknown I am not sure how you able to solve the above without using, conservation of momentum and conservation of energy (elastic collision). The combination of the two yields 2 equation and 2 unknown which you then solve: Conservation of "Momentum": #m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2# =======> (1) Let, #m_1 = 2kg; m_2 = 1 kg; v_1=9m/s; v_2=0m/s# Conservation of energy (elastic collision): #1/2m_1v_1^2 + 1/2m_2v_2^2 = 1/2m_1v'_1^2 + 1/2m_2v'_2^2 # =======> (2) We have 2 equations and 2 unknowns: From (1) ==> #2*9 = 2v'_1 + v'_2; color(blue)(v'_2 = 2(9-v'_1))# ==>(3) From (2) ==> #9^2 = v'_1^2 + 1/2v'_2^2# ===================> (4) Insert # (3) => (4)#: #9^2 = v'_1^2 + 1/2*[color(blue)[2(9-v'_1)]]^2# expand #9^2 = v'_1^2 + 2(9^2-18v'_1 + v'_1^2)# #2v'_1^2 -36v'_1 + 9^2 = 0# solve the quadratic equation for #v'_1# Using the quadratic formula: #v'_1 = (b +-sqrt(b^2 - 4ac)/2a); v'_1 => (2.64, 15.36) # The solution that make sense is 2.64 (explain why?) Insert in (3) and solve #color(blue)(v'_2 = 2(9-color(red)2.64) = 12.72# So the speed after collision of the two objects are: #v'_1 = 2.64 m/s, v'_2 = 12.72#

William Audy · Answer

#v_1=3 m/s#
#v_2=12 m/2# #m_1*v_1+m_2*v_2=m_1*v_1'+m_2*v_2^'" (1)"# #cancel(1/2)*m_1*v_1^2+cancel(1/2)*m_2*v_2^2=cancel(1/2)*m_1*v_1^('2)+cancel(1/2)*m_2*v_2^('2) "# #m_1*v_1^2+m_2*v_2^2=m_1*v_1^('2)+m_2*v_2^('2)" (2)"# #m_1*v_1-m_1*v_1^'=m_2*v_2^'-m_2*v_2" redeployment of (1)"# #m_1(v_1-v_1^')=m_2(v_2^'-v_2)" (3)"# #m_1*v_1^2-m_1*v_1^('2)=m_2*v_2^('2)-m_2*v_2^2" redeployment of (2)"# #m_1(v_1^2-v_1^('2))=m_2(v_2^('2)-v_2^2)" (4)"# #"divide :(3)/(4)"# #(m_1(v_1-v_1^'))/(m_1(v_1^2-v_1^('2)))=(m_2(v_2^'-v_2))/(m_2(v_2^('2)-v_2^2))# #(v_1-v_1^')/((v_1^2-v_1^('2)))=((v_2^'-v_2))/((v_2^('2)-v_2^2))# #v_1^2-v_1^('2)=(v_1+v_1^')*(v_1-v_1^') ; v_2^('2)=(v_2^'+v_2)*(v_2^'-v_2)# #v_1+v_1^'=v_2+v_2^'#

Leo Abeyta · Answer

undefined The post-collision velocities of the 2 kg ball and the 1 kg ball are 6 m/s and 12 m/s, respectively.