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A 0.142 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.740m/s . It has a head-on collision with a 0.310 kg glider that is moving to the ...show more
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As the collision is elastic, both momentum and kinetic energy remain conserved

Suppose velocity towards right is taken as positive then velocity to the left will be negative

Mass of lighter glider =m1=0.142 kg

vel.of m1 before collision u1=0.740 m/s to the right

Mass of heavier glider = m2 = 0.310 kg

vel.of m2 before collision u2= - 2.29 m/s (negative because it is to left)

If final vel. of m1 is v1 and of mass m2 is v2, then

as momentum is conserved,

m1u1 + m2u2 = m1v1 +m2v2_____________(1)

as kinetic energy is conserved,

(1/2)m1u1^2+(1/2)m2u2^2 = (1/2)m1v1^2+(1/2)m2v2^2__(2)

solving these equations,

v1=[2 m2u2 +(m1-m2) u1] / (m1+m2)

v2 = [2 m1u1 +(m2-m1) u2] / (m1+m2)

substituting vlues of u1,u2 etc we get

v1 = { 2*0.310*( -2.29) + [( 0.142 -0.310)0.74] } / [ 0.142+0.310]

v1 =-1544.12/452= - 3.416m/s
A) the magnitude of the final velocity of the 0.142kg glider.
v1= - 2.86m/s

B)velocity is to left because it is negative

C) the magnitude of the final velocity of the 0.310 kg glider.

velocity is negative, it is v2=0.386 m/s

D) velocity of0.310 kg is negative , it is to left

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