Two blocks move along a linear path on a nearly frictionless air track. One bloc

Question

Two blocks move along a linear path on a nearly frictionless air track. One block, of mass 0.125 kg, initially moves to the right at a speed of 5.10 m/s, while the second block, of mass 0.250 kg, is initially to the left of the first block and moving to the right at 6.70 m/s. Find the final velocities of the blocks, assuming the collision is elastic.

THE ANSWERS FOR THIS SAMPLE PROBLEM ARE 7.23 M/S AND 5.63 M/S. HOW DO I FIND THESE ANSWERS. i HAVE THE EQUATIONS BUT NONE OF MY ANSWERS CAME CLOSE TO 7.23 AND 5.63. CAN YOU HELP ME. EXPLAIN YOUR ANSWER TOO PLEASE

Explanation / Answer

as you figured it out we have two equations

1. conservation of momentum

2. conservation of energy

apply first equation

m1v1i+m2v2i = m1v1f + m2v2f

as we know m1= 0.125 kg v1i = 5.1 m/s taking right side as positive *(dont forget to take signs for velocity as they are vectors and direction dependent)

m2 = 0.25 kg v2i = -6.7 m/s

so we have

0.125*5.1+0.25*(-6.7) = 0.125 v1f+0.25v2f

v1f+2*v2f = -8.3

second equation is conservation of energy as the collision is elastic no loss in collision

so

1/2*m1v1i2+ 1/2*m2v2i2 = 1/2*m1v1f2+1/2*m2v2f2

so 0.125*(5.1)2+0.25(-6.7)2 = 0.125*v1f2+0.25*v2f2

v1f2+2*v2f2 = 115.79

so now we do have two equations

now solve for v1f and v2f

(8.3+2*v2f)2+2*v2f2 = 115.79

v2f = 1.17 or -6.7

as the collision takes place higher momentum towards left hand side so both of them move left

so v2f = -6.7m/s

v1f = 5.1 m/s

my answers came like this

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