A 320-g air track cart traveling at 1.25 m/s suddenly collides elastically with a stationary 270-g cart. What is the speed of 320-g cart just after the collision?

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Answer 1 · 2018-07-26T11:40:45

The speed of the first cart is #=0.106ms^-1#

There is conservation of momentum and conservation of kinetic energy as the collision is elastic

#m_1u_1+m_2u_2=m_1v_1+m_2v_2#

And

#1/2m_1u_1^2+1/2m_2u_2^2=1/2m_1v_1^2+1/2m_2v_2^2#

Plugging in the above values

#0.32*1.25+0.27*0=0.32v_1+0.27v_2#

#0.32v_1+0.27v_2=0.4#.............................#(1)#

#1/2*0.32*1.25^2+1/2*0.27*0=1/2*0.32*v_1^2+1/2*0.27*v_2^2#

#0.32v_1^2+0.27v_2^2=0.5#....................#(2)#

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

#0.32v_1^2+0.27(0.4-0.32v_1)^2/0.27^2=0.5#

#0.0864v_1^2+0.16+0.1024v_1^2-0.256v_1=0.135#

#0.1888v_1^2-0.256v_1+0.025=0#

#v_1=(0.256+-sqrt(0.256^2-4*0.1888*0.025))/(2*0.188)#

#v_1=(0.256+-0.216)/(0.3776)#

Therefore,

#v_1=1.25ms^-1# or #v_1=0.106ms^-1#

And

#v_2=0# or #v_2=1.36ms^-1#

Discard #v_1=1.25# and #v_2=0# as those are the speeds before collision.