A 0.150kg glider is moving to the right (+x) on a frictionless, horizontal air t

Question

A 0.150kg glider is moving to the right (+x) on a frictionless, horizontal air track with a speed of 0.80m/s. It has an elastic collision with a 0.300kg glider moving to the left (-x) with a speed of 2.20m/s. a.) What is the initial momentum of each glider? Express the momentum in terms of unit vectors. b.) Use the relative velocity formula (Equation 8.21) to find v2f in terms of v1f. c.) Use the relative velocity result to solve conservation of momentum to find the velocity (magnitude and direction) of each glider after the collision. d.) Show that kinetic energy is conserved.

Explanation / Answer

momentum = mass x velocity a) Let glider 1 be the glider moving to the right.Let glider 2 be the glider moving to the left. the vector i denotes direction in the x direction initial momentum of glider 1= (0.150)(0.80)=0.12 i Ns initial momentum of glider 2=0.300)(2.20)= -0.66i Ns b) using the concept of conservation of energy and conservation of momentum. pi=pf m1u1+m2u2=m1v1+m2v2 or -0.54=0.150v1+0.300v2 or v2=(-0.54-0.15v1)/(0.300) d) to show KE is conserved, find KEi and also KEf then compare the results.

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